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176. Given the functions $f(x)=x^{2}-x$ and $g(x)=x^{2}+x$. It is required to determine: a) at what value of the argument the function $f(x)$ increases twice as fast as $g(x) ; \sigma)$ does there exist a value of $x$ at which the functions change at the same rate? | Solution. a) We find the derivatives of the given functions: $f^{\prime}(x)=$ $=2 x-1$ and $g^{\prime}(x)=2 x+1$ (see solutions of examples 172 and 173). According to the condition, the equality $2 x-1=2(2 x+1)$ must hold; solving this equation, we find $x=-3 / 2$.
b) Since the equation $2 x-1=2 x+1$ has no real roots... | -\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,965 |
177. Find the derivative of the function $y=2 x^{3}-x^{2}+5$ at the point $x=2$. | Solution. First, we find the derivative of the given function in general form, i.e., at an arbitrary point $x$:
$1^{0} . y_{\mathrm{H}}=y+\Delta y=2(x+\Delta x)^{3}-(x+\Delta x)^{2}+5=2 x^{3}+6 x^{2} \Delta x+6 x(\Delta x)^{2}+$ $+2(\Delta x)^{3}-x^{2}-2 x \Delta x-(\Delta x)^{2}+5$
$2^{0} . \Delta y=6 x^{2} \Delta x... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,966 | |
183. Given the function $y=\sqrt{x}$. Find $y_{x=9}^{\prime}$. | Solution. $1^{0} . y_{n}=\sqrt{x+\Delta x}$.
$2^{0} . \Delta y=y_{\mathrm{n}}-y=\sqrt{x \pm \Delta x}-\sqrt{x}$.
$3^{0} . \frac{\Delta y}{\Delta x}=\frac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,967 | |
186. Using the definition of the derivative, find the derivative of the function $y=\sin 3x$ at the point $x=\pi / 9$. | Solution. Let's give some value $x$ an increment $\Delta x$; then the function $y$ will receive an increment
$$
\Delta y=\sin (3(x+\Delta x))-\sin 3 x=2 \cos \left(3 x+\frac{3 \Delta x}{2}\right) \sin \frac{3 \Delta x}{2}
$$
Form the ratio $\frac{\Delta y}{\Delta x}$ and take the limit as $\Delta x \rightarrow 0$:
$... | \frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,968 |
189. Show that the function $y=\tan x$ does not have a derivative at the point $x=\pi / 2$.
untranslated portion:
```
Показать, что в точке $x=\pi / 2$ функция $y=\operatorname{tg} x$ не имеет производной.
```
Here is the translated text with the original formatting preserved:
189. Show that the function $y=\tan x... | Solution. At the point $x=\pi / 2$ the function $\operatorname{tg} x$ does not exist, i.e., the condition of continuity of the function is not satisfied; therefore, the function does not have a derivative at this point. | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,969 |
192. Find the derivative of the function $y=x^{2}$.
翻译完成，保留了原文的格式和换行。 | Solution. Let's represent the given function as a product: $x^{2}=x \cdot x$. Now, using rule IV, we find
$$
\left(x^{2}\right)^{\prime}=(x \cdot x)^{\prime}=x^{\prime} x+x x^{\prime}=2 x
$$
since $(x)^{\prime}=1$. Therefore, $\left(x^{2}\right)^{\prime}=2 x$. | 2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,970 |
194. Find the derivative of the function $y=x^{3}$.
翻译完成，保留了原文的格式和换行。 | Solution. Representing the given function as a product: $x^{3}=x^{2} \cdot x$ and applying rule IV again, we find
$$
\left(x^{3}\right)^{\prime}=\left(x^{2} \cdot x\right)^{\prime}=\left(x^{2}\right)^{\prime} \cdot x+(x)^{\prime} \cdot x^{2}=2 x \cdot x+1 \cdot x^{2}=2 x^{2}+x^{2}=3 x^{2}
$$ | 3x^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,971 |
197. Find the derivative of the function $y=9 x^{5}$. | Solution. Using rule V and the formula $\left(x^{n}\right)^{\prime}=n x^{n-1}$, we get
$$
y^{\prime}=\left(9 x^{5}\right)^{\prime}=9 \cdot 5 x^{4}=45 x^{4}
$$
When skilled, intermediate steps can be skipped:
$$
\left(9 x^{5}\right)^{\prime}=45 x^{4}
$$ | 45x^{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,972 |
199. Find the derivative of the function $y=5 x^{2}-x+4$. | Solution. $\left(5 x^{2}-x+4\right)^{\prime}=\left(5 x^{2}\right)^{\prime}-(x)^{\prime}+(4)^{\prime}=5\left(x^{2}\right)^{\prime}-1=10 x-1$. 200-214. Find the derivatives of the following functions:
^{\prime}=\left(x^{-3}\right)^{\prime}=-3 x^{-3-1}=-3 x^{-4}=-3 \cdot \frac{1}{x^{4}}=-\frac{3}{x^{4}}
$$ | -\frac{3}{x^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,974 |
220. Differentiate the function $y=2 x^{3}\left(x^{6}-1\right)$. | Solution. Method I. Using rule IV, we get
$$
\begin{aligned}
& \left(2 x^{3}\left(x^{6}-1\right)\right)^{\prime}=\left(2 x^{3}\right)^{\prime}\left(x^{6}-1\right)+\left(x^{6}-1\right)^{\prime} \cdot 2 x^{3}=6 x^{2}\left(x^{6}-1\right)+6 x^{5} \cdot 2 x^{3}= \\
& =6 x^{8}-6 x^{2}+12 x^{8}=18 x^{8}-6 x^{2}=6 x^{2}\left(... | 6x^{2}(3x^{6}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,975 |
221. Differentiate the function $y=\left(x^{2}+2\right)(2 x+1)$. | Solution. Method I. We will use Rule IV with $u=$ $=x^{2}+2, v=2 x+1$. We have
$$
\begin{gathered}
y^{\prime}=\left(x^{2}+2\right)^{\prime}(2 x+1)+(2 x+1)^{\prime}\left(x^{2}+2\right)=2 x(2 x+1)+2\left(x^{2}+2\right)= \\
=4 x^{2}+2 x+2 x^{2}+4=6 x^{2}+2 x+4
\end{gathered}
$$
Method II. First, we multiply the expressi... | 6x^2+2x+4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,976 |
228. Find the derivative of the function $y=\frac{x^{2}-2}{x^{2}+2}$. | Solution. Applying rule VI:
$$
\begin{gathered}
y^{\prime}=\frac{\left(x^{2}-2\right)^{\prime}\left(x^{2}+2\right)-\left(x^{2}+2\right)^{\prime}\left(x^{2}-2\right)}{\left(x^{2}+2\right)^{2}}=\frac{2 x\left(x^{2}+2\right)-2 x\left(x^{2}-2\right)}{\left(x^{2}+2\right)^{2}}= \\
=\frac{2 x\left(x^{2}+2-x^{2}+2\right)}{\l... | \frac{8x}{(x^{2}+2)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,977 |
229. Find the derivative of the function $y=\frac{1}{x}$. | Solution. 1 using the rule. We apply rule V1 with $\boldsymbol{u}=1, \boldsymbol{v}=\boldsymbol{x}$. Then we get
$$
y^{\prime}=\frac{0 \cdot x-1 \cdot 1}{x^{2}}=-\frac{1}{x^{2}}
$$
And the method. We first transform the given function to the form $y=\frac{1}{x}=x^{-1}$, and then apply the formula $\left(x^{n}\right)=... | -\frac{1}{x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,978 |
239. Given that $(\sin t)^{\prime}=\cos t$, differentiate the function $y=\sin 2 x$. | Solution. The complexity coefficient of the given function is 2. The order of the intermediate functions is as follows: $y=\sin u, u=2 x$. We find
$$
y^{\prime}=\cos 2 x \cdot(2 x)^{\prime}=\cos 2 x \cdot 2=2 \cos 2 x
$$ | 2\cos2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,979 |
240. Find $y^{\prime}$, if $y=\sin ^{2} 4 x$. | Solution. We have $y=\sin ^{2} 4 x$ (3). The order of functions: $y=u^{2} ; u=\sin v ; v=4 x$. Therefore,
$y^{\prime}=2 \sin 4 x(\sin 4 x)^{\prime}=2 \sin 4 x \cos 4 x \cdot(4 x)^{\prime}=2 \sin 4 x \cos 4 x \cdot 4=4 \sin 8 x$ (in the final expression, the double-angle sine formula is used).
In the future, when suff... | 4\sin8x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,980 |
249. $y=\ln x^{2}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
249. $y=\ln x^{2}$. | Solution. $y^{\prime}=\frac{1}{x^{2}} \cdot\left(x^{2}\right)^{\prime}=\frac{1}{x^{2}} \cdot 2 x=\frac{2}{x}$. | \frac{2}{x} | Calculus | proof | Yes | Yes | olympiads | false | 31,981 |
250. $y=\ln \left(x^{3}-1\right)$.
250. $y=\ln \left(x^{3}-1\right)$.
The above text has been translated into English, retaining the original text's line breaks and format. However, since the original text is a mathematical expression, the translation is identical to the original as mathematical expressions are gene... | Solution. $y^{\prime}=\frac{1}{x^{3}-1}\left(x^{3}-1\right)^{\prime}=\frac{1}{x^{3}-1} \cdot 3 x^{2}=\frac{3 x^{2}}{x^{3}-1}$. | \frac{3x^{2}}{x^{3}-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,982 |
253. $y=\ln \sin x$. | Solution. Here $u=\sin x$. Then we get
$$
y_{x}^{\prime}=\frac{u_{x}^{\prime}}{u}=\frac{(\sin x)^{\prime}}{\sin x}=\frac{\cos x}{\sin x}=\operatorname{ctg} x
$$ | \operatorname{ctg}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,983 |
254. $f(x)=\ln ^{3}\left(x^{2}-1\right)$.
254. $f(x)=\ln ^{3}\left(x^{2}-1\right)$.
The translation is the same as the original text because it is a mathematical expression which is universal and does not change when translated into English. However, if you meant to have the problem number and the function descripti... | Solution. We have $f(x)=\ln ^{3}\left(x^{2}-1\right)$. The order of intermediate functions is as follows: $f(x)=u^{3} ; u=\ln v ; v=x^{2}-1$. Therefore,
$$
f^{\prime}(x)=3 \ln ^{2}\left(x^{2}-1\right) \cdot \frac{1}{x^{2}-1} \cdot 2 x=\frac{6 x}{x^{2}-1} \ln ^{2}\left(x^{2}-1\right) .
$$ | \frac{6x}{x^{2}-1}\ln^{2}(x^{2}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,984 |
264. $y=\log _{3} 4 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
264. $y=\log _{3} 4 x$. | Solution. Method I. Since $\log _{3} 4 x=\log _{3} 4+\log _{3} x$, then
$$
y^{\prime}=\left(\log _{3} 4+\log _{3} x\right)^{\prime}=\left(\log _{3} 4\right)^{\prime}+\left(\log _{3} x\right)^{\prime}=0+\frac{1}{x \ln 3}=\frac{1}{x \ln 3}
$$
Method II. $\left(\log _{3} 4 x\right)^{\prime}=\frac{1}{4 x \ln 3}(4 x)^{\pr... | \frac{1}{x\ln3} | Algebra | proof | Yes | Yes | olympiads | false | 31,985 |
267. $u=\log _{7} x^{5}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
267. $u=\log _{7} x^{5}$. | Solution. I method. $\left(\log _{7} x^{5}\right)^{\prime}=\left(5 \log _{7} x\right)^{\prime}=\frac{5}{x \ln 7}$.
II method. $\left(\log _{7} x^{5}\right)^{\prime}=\frac{1}{x^{5} \ln 7}\left(x^{5}\right)^{\prime}=\frac{5 x^{4}}{x^{5} \ln 7}=\frac{5}{x \ln 7}$. | Algebra | proof | Yes | Yes | olympiads | false | 31,986 | |
270. $y=\lg (x-1)$.
270. $y=\lg (x-1)$.
The translation is provided while retaining the original text's line breaks and format. However, since the original text is a mathematical expression, the translation is identical to the source as it is already in a universally recognized form. If you intended for a descriptio... | Solution. $y^{\prime}=\frac{1}{(x-1) \ln 10}$. | \frac{1}{(x-1)\ln10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,987 |
273. $y=\log _{3}\left(x^{2}+3 x-1\right)$.
273. $y=\log _{3}\left(x^{2}+3 x-1\right)$. | Solution. $y^{\prime}=\frac{\left(x^{2}+3 x-1\right)^{\prime}}{\left(x^{2}+3 x-1\right) \ln 3}=\frac{2 x+3}{\left(x^{2}+3 x-1\right) \ln 3}$. | \frac{2x+3}{(x^{2}+3x-1)\ln3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,988 |
276. $u=\log _{5} \cos 7 x$.
276. $u=\log _{5} \cos 7 x$. | Solution. $u^{\prime}=\frac{1}{\cos 7 x \ln 5}(\cos 7 x)^{\prime}=\frac{-\sin 7 x \cdot 7}{\cos 7 x \ln 5}=-\frac{7 \operatorname{tg} 7 x}{\ln 5}$. | -\frac{7\operatorname{tg}7x}{\ln5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,989 |
279. $y=2 x^{3}-4 x+2$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
279. $y=2 x^{3}-4 x+2$. | Solution. $y^{\prime}=\left(2 x^{3}-4 x+2\right)^{\prime}=\left(2 x^{3}\right)^{\prime}-(4 x)^{\prime}+(2)^{\prime}=6 x^{2}-4$. | 6x^{2}-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,990 |
280. $y=\frac{3}{5 x^{2}}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
280. $y=\frac{3}{5 x^{2}}$. | Solution. Let's transform the given function to the form $y=\frac{3}{5 x^{2}}=$ $=\frac{3}{5} x^{-2}$. Then we get
$$
y^{\prime}=\left(\frac{3}{5} x^{-2}\right)^{\prime}=\frac{3}{5}\left(x^{-2}\right)^{\prime}=\frac{3}{5}(-2) x^{-3}=-\frac{6}{5} x^{-3}=-\frac{6}{5 x^{3}}
$$ | Algebra | proof | Yes | Yes | olympiads | false | 31,991 | |
285. $f(x)=\sqrt[4]{x^{3}}$.
285. $f(x)=\sqrt[4]{x^{3}}$.
The translation is the same as the original text because it is a mathematical expression which is universal and does not change when translated into English. However, if you need a description or explanation of the function in English, please let me know. | Solution. Since $f(x)=\sqrt[4]{x^{3}}=x^{3 / 4}$, then
$$
f^{\prime}(x)=\left(x^{3 / 4}\right)^{\prime}=\frac{3}{4} x^{3 / 4-1}=\frac{3}{4} x^{-1 / 4}=\frac{3}{4 \sqrt[4]{x}}
$$ | \frac{3}{4\sqrt[4]{x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,992 |
286. $y=\frac{2}{3 x \sqrt{x}}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
286. $y=\frac{2}{3 x \sqrt{x}}$. | Solution. Here $y=\frac{2}{3 x \sqrt{x}}=\frac{2}{3} x^{-3 / 2}$, hence
$$
y^{\prime}=\left(\frac{2}{3} x^{-3 / 2}\right)^{\prime}=-\frac{3}{2} \cdot \frac{2}{3} x^{-5 / 2}=-x^{-5 / 2}=-\frac{1}{x^{2} \sqrt{x}}
$$ | -\frac{1}{x^{2}\sqrt{x}} | Algebra | proof | Yes | Yes | olympiads | false | 31,993 |
287. $y=x^{2}+\frac{2}{x^{4}}-\sqrt[3]{x}$
287. $y=x^{2}+\frac{2}{x^{4}}-\sqrt[3]{x}$ | Solution. We have $y=x^{2}+2 x^{-4}-x^{1 / 3}$. Therefore,
$$
\begin{aligned}
y^{\prime}=\left(x^{2}+2 x^{-4}-x^{1 / 3}\right)^{\prime}= & \left(x^{2}\right)^{\prime}+\left(2 x^{-4}\right)^{\prime}-\left(x^{1 / 3}\right)^{\prime}=2 x-8 x^{-5}-\frac{1}{3} x^{-2 / 3}= \\
& =2 x-\frac{8}{x^{5}}-\frac{1}{3 \sqrt[3]{x^{2}}... | 2x-\frac{8}{x^{5}}-\frac{1}{3\sqrt[3]{x^{2}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,994 |
288. $y=\frac{3 x^{2}-4 x+1}{2 x+1}$.
288. $y=\frac{3 x^{2}-4 x+1}{2 x+1}$. | Solution. We apply the quotient rule of differentiation and formula $\mathrm{X}$ :
$$
\begin{gathered}
y^{\prime}=\frac{\left(3 x^{2}-4 x+1\right)^{\prime}(2 x+1)-(2 x+1)^{\prime}\left(3 x^{2}-4 x+1\right)}{(2 x+1)^{2}}= \\
=\frac{(6 x-4)(2 x+1)-2\left(3 x^{2}-4 x+1\right)}{(2 x+1)^{2}}=\frac{12 x^{2}+6 x-8 x-4-6 x^{2... | \frac{6(x^{2}+x-1)}{(2x+1)^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,995 |
289. $y=\sqrt{x}\left(x^{2}+2 x-5\right)$.
289. $y=\sqrt{x}\left(x^{2}+2 x-5\right)$. | Solution. Here we should use the product differentiation rule and formula X. We have
$$
y^{\prime}=(\sqrt{x})^{\prime}\left(x^{2}+2 x-5\right)+\left(x^{2}+2 x-5\right)^{\prime} \sqrt{x}
$$
Since $(\sqrt{x})^{\prime}=\left(x^{1 / 2}\right)^{\prime}=\frac{1}{2} x^{-1 / 2}=\frac{1}{2 \sqrt{x}},\left(x^{2}+2 x-5\right)^{... | 2.5x\sqrt{x}+3\sqrt{x}-\frac{5}{2\sqrt{x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,996 |
296. $f(x)=x^{3} \sin x$.
296. $f(x)=x^{3} \sin x$.
The translation is provided while retaining the original format and line breaks. However, since the original text is a mathematical expression, the translation is identical to the source text as it is already in English. | Solution. Applying formulas IV, X and considering that $(\sin x)^{\prime}=$ $=\cos x$, we find
$f^{\prime}(x)=\left(x^{3}\right)^{\prime} \sin x+(\sin x)^{\prime} x^{3}=3 x^{2} \sin x+x^{3} \cos x=x^{2}(3 \sin x+x \cos x)$. | x^{2}(3\sinx+x\cosx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,997 |
301. Find the derivative of the function $y=5^{x}$. | Solution. Using the formula $\left(a^{x}\right)^{\prime}=a^{x} \ln a$, we get
$$
y^{\prime}=\left(5^{x}\right)^{\prime}=5^{x} \ln 5
$$ | y^{\}=5^{x}\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,998 |
302. Differentiate the function $y=2^{\sin x}$. | Solution. $y^{\prime}=2^{\sin x} \ln 2(\sin x)^{\prime}=2^{\sin x} \ln 2 \cos x$. | 2^{\sinx}\ln2\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,999 |
303. Find the derivative of the function $y=0.5^{\sin 4 x}$. | Solution. $\quad y^{\prime}=0.5^{\sin 4 x} \ln 0.5(\sin 4 x)^{\prime}=0.5^{5 \sin 4 x} \ln 0.5 \cdot \cos 4 x \cdot 4=$ $=4 \cos 4 x \cdot 0.5^{\sin 4 x} \ln 0.5$. | 4\cos4x\cdot0.5^{\sin4x}\ln0.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,000 |
304. Differentiate the function $y=e^{x^{2}}$. | Solution. We use the formula ( $\left.e^{u}\right)^{\prime}=e^{u} u^{\prime}$ :
 | y'=2xe^{x^2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,001 |
305. Find the value of the derivative of the function $y=2^{x}$ at the point $x=1$. | Solution. We find the derivative using formula XI: $y^{\prime}=2^{x} \ln 2$. We compute the particular value of the derivative at $x=1$:
$$
y^{\prime}(1)=\left(2^{x} \ln 2\right)_{x=1}=2 \ln 2=\ln 4 .
$$
306-311. Find the derivatives of the following functions: | \ln4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,002 |
314. Find the derivative of the function $y=\sin \left(x^{3}-3 x^{2}\right)$. | Solution. Here $u=x^{3}-3 x^{2}, y=\sin u$. We find $u_{x}^{\prime}=\left(x^{3}-\right.$ $\left.-3 x^{2}\right)^{\prime}=3 x^{2}-6 x$. Then $y_{x}^{\prime}=y_{u}^{\prime} u_{x}^{\prime}=\left(3 x^{2}-6 x\right) \cos \left(x^{3}-3 x^{2}\right)$. | (3x^{2}-6x)\cos(x^{3}-3x^{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,003 |
317. Find $y^{\prime}$, if $y=\ln \sin ^{3} 5 x$. | Solution. We will use formulas VII, VIII, and XIII:
$$
y^{\prime}=\frac{1}{\sin ^{3} 5 x} \cdot 3 \sin ^{2} 5 x \cos 5 x \cdot 5=\frac{15 \cos 5 x}{\sin 5 x}=15 \operatorname{ctg} 5 x
$$
318-323. Find the derivatives of the following functions: | 15\operatorname{ctg}5x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,004 |
326. Find the derivative of the function $y=\cos x^{4}$. | Solution. Here $u=x^{4}, y=\cos u$. We find $y_{u}^{\prime}=-\sin u, u_{x}^{\prime}=$ $=4 x^{3}$. Therefore, $y_{x}^{\prime}=y_{u}^{\prime} u_{x}^{\prime}=-4 x^{3} \sin x^{4}$. | -4x^{3}\sinx^{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,005 |
327. Find $\left(\cos ^{3} x\right)^{\prime}$. | Solution. Here $u=\cos x, y=u^{3}$. We find $y_{u}^{\prime}=3 u^{2}=3 \cos ^{2} x$, $u_{x}^{\prime}=(\cos x)^{\prime}=-\sin x$; hence $y_{x}^{\prime}=3 \cos ^{2} x(-\sin x)=-3 \cos ^{2} x \sin x$. | -3\cos^{2}x\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,006 |
328. Find the derivative of the function $y=\cos \left(\frac{x}{2}-x^{5}\right)$. | Solution. We have $u=\frac{x}{2}-x^{5}, y=\cos u$. Further, we find $y_{u}^{\prime}=$ $=-\sin u, u_{x}^{\prime}=\frac{1}{2}-5 x^{4}$, from which $y_{x}^{\prime}=-\left(\frac{1}{2}-5 x^{4}\right) \sin \left(\frac{x}{2}-x^{5}\right)$. | y_{x}^{\}=-(\frac{1}{2}-5x^{4})\sin(\frac{x}{2}-x^{5}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,007 |
329. Find the value of the arbitrary function $f(x)=\sin ^{4} x-$ $-\cos ^{4} x$ at $x=\pi / 12$. | Solution. First, we transform the given function:
$$
f(x)=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)=-\cos 2 x
$$
(since $\sin ^{2} \alpha+\cos ^{2} \alpha=1, \cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha$). Using the rule for differentiating a composite function, we get
$$
f^{\prime}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,008 |
330. Find $y^{\prime}$, if $y=\sqrt{x}+\cos ^{2} 3 x$. | Solution. $\quad y^{\prime}=\left(\sqrt{x}+\cos ^{2} 3 x\right)^{\prime}=(\sqrt{x})^{\prime}+\left(\cos ^{2} 3 x\right)^{\prime}=\frac{1}{2 \sqrt{x}}+$ $+2 \cos 3 x(-\sin 3 x) \cdot 3=\frac{1}{2 \sqrt{x}}-3 \sin 6 x$.
Here we used the fact that $(\sqrt{x})^{\prime}=\left(x^{1 / 2}\right)^{\prime}=\frac{1}{2} x^{-1 / 2... | \frac{1}{2\sqrt{x}}-3\sin6x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,009 |
338. Find the derivative of the function $y=\operatorname{tg} x^{5}$. | Solution. Here $u=x^{5}, y=\operatorname{tg} u$. Therefore,
$$
\left(\operatorname{tg} x^{5}\right)^{\prime}=\frac{1}{\cos ^{2} x^{5}} \cdot\left(x^{5}\right)^{\prime}=\frac{5 x^{4}}{\cos ^{2} x^{5}}
$$ | \frac{5x^{4}}{\cos^{2}x^{5}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,010 |
339. Find the derivative of the function $y=\operatorname{ctg}^{3} x$. | Solution. $y^{\prime}=\left(\operatorname{ctg}^{3} x\right)^{\prime}=3 \operatorname{ctg}^{2} x\left(-\frac{1}{\sin ^{2} x}\right)=-\frac{3 \operatorname{ctg}^{2} x}{\sin ^{2} x}$. | -\frac{3\operatorname{ctg}^{2}x}{\sin^{2}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,011 |
343. Find $y^{\prime}$, if $y=\arcsin x^{3}$. | solution. Using the formula $(\arcsin u)^{\prime}=\frac{u^{\prime}}{\sqrt{1-u^{2}}}$, we have
$$
y^{\prime}=\left(\arcsin x^{3}\right)^{\prime}=\frac{\left(x^{3}\right)^{\prime}}{\sqrt{1-\left(x^{3}\right)^{2}}}=\frac{3 x^{2}}{\sqrt{1-x^{6}}}
$$ | \frac{3x^{2}}{\sqrt{1-x^{6}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,012 |
344. Find the derivative of the function $y=\arcsin \ln x$. | Solution. We again apply the formula $(\arcsin u)^{\prime}=\frac{u^{\prime}}{\sqrt{1-u^{2}}}$ :
$$
y^{\prime}=(\arcsin \ln x)^{\prime}=\frac{(\ln x)^{\prime}}{\sqrt{1-(\ln x)^{2}}}=\frac{1}{x \sqrt{1-\ln ^{2} x}} .
$$ | \frac{1}{x\sqrt{1-\ln^{2}x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,013 |
354. Find $y^{\prime}$, if $y=\operatorname{arctg} 2 x$. | Solution. $y^{\prime}=\frac{1}{1+(2 x)^{2}} \cdot(2 x)^{\prime}=\frac{2}{1+4 x^{2}}$. | \frac{2}{1+4x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,014 |
355. Find $y^{\prime}$, if $y=\operatorname{arcctg} 3 x$. | Solution. $y^{\prime}=-\frac{1}{1+(3 x)^{2}} \cdot(3 x)^{\prime}=-\frac{3}{1+9 x^{2}}$. | -\frac{3}{1+9x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,015 |
356. Find $y^{\prime}$, if $y=\operatorname{arctg} x^{2}$. | Solution. $y^{\prime}=\frac{1}{1+x^{4}} \cdot\left(x^{2}\right)^{\prime}=\frac{2 x}{1+x^{4}}$. | \frac{2x}{1+x^{4}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,016 |
357. Find \( f^{\prime}(1 / 5) \) if \( f(x)=\operatorname{arctg} 5 x+x^{2} \). | Solution. We have
$$
f^{\prime}(x)=\frac{1}{1+(5 x)^{2}} \cdot(5 x)^{\prime}+2 x=\frac{5}{1+25 x^{2}}+2 x
$$
Therefore,
$$
f^{\prime}(1 / 5)=\frac{5}{1+25 \cdot \frac{1}{25}}+2 \cdot \frac{1}{5}=\frac{5}{2}+\frac{2}{5}=2.5+0.4=2.9
$$ | 2.9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,017 |
358. Given: $f(x)=\operatorname{arcctg} \frac{x}{2}-\operatorname{arctg} \sqrt{x}$. Find $f^{\prime}(1)$. | Solution. $f^{\prime}(x)=-\frac{1}{1+\left(\frac{x}{2}\right)^{2}} \cdot\left(\frac{x}{2}\right)^{\prime}-\frac{1}{1+(\sqrt{x})^{2}} \cdot(\sqrt{x})^{\prime}=$
$$
\begin{aligned}
& =-\frac{\frac{1}{2}}{1+\frac{x^{2}}{4}}-\frac{1}{1+x} \cdot \frac{1}{2 \sqrt{x}}=-\frac{1}{2\left(1+\frac{x^{2}}{4}\right)}-\frac{1}{2(1+x... | -\frac{13}{20} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,018 |
373. At what point is the tangent to the curve $y=x^{2}-1$ : a) parallel to the $O x$ axis; b) forms a $45^{\circ}$ angle with the $O x$ axis? | Solution. a) Since the line is parallel to the $O x$ axis, it forms an angle of 0 with it, and its angular coefficient, equal to the tangent of this angle, is zero. It is known that the derivative of the function at a given point is equal to the angular coefficient of the tangent to the curve at this point, i.e., $f^{\... | (0,-1)(\frac{1}{2},-\frac{3}{4}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,019 |
374. Find the slope of the tangent line drawn to the curve $y=x^{3}$ at the point $C(-2;-8)$. | Solution. Let's find the derivative of the function $y=x^{3}$ at the point $x=-2$:
$$
y^{\prime}=\left(x^{3}\right)^{\prime}=3 x^{2} ; \quad y_{x=-2}^{\prime}=3(-2)^{2}=12
$$
Thus, the slo... | 12 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,020 |
375. The curve is given by the equation $y=x^{2}+5 x+3$. Determine the angles of inclination of the tangents to the positive direction of the $O x$ axis, drawn to the curve at the points with abscissas $x=-2$ and $x=0$. | Solution. Let's find the derivative: $y^{\prime}=2 x+5$. Denoting the angle of inclination of the tangent at the point with abscissa $x=-2$ by $\alpha$, and at the point with abscissa $x=0$ by $\beta$, we get
$$
\operatorname{tg} \alpha=y_{x=-2}^{\prime}=2(-2)+5=1, \operatorname{tg} \beta=y_{x=0}^{\prime}=2 \cdot 0+5=... | \alpha=45,\beta\approx79 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,021 |
376. On the curve $y=4 x^{2}-6 x+3$, find the point where the tangent line is: a) parallel to the line $y=2 x$; b) perpendicular to the line $y=x / 4$. | Solution. Let the desired point of tangency be ( $x_{0} ; y_{0}$ ). Then, as is known, the slope $k$ of the tangent line is equal to the value of the derivative at the point of tangency, i.e.,
$$
k=y_{x=x_{0}}^{\prime}=8 x-\left.6\right|_{x=x_{0}}=8 x_{0}-6=2\left(4 x_{0}-3\right)
$$
Considering this, let's examine e... | (1;1)(1/4;7/4) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,022 |
377. Find the angles at which the parabola $y=x^{2}-x$ intersects the x-axis (Fig. 105). | Solution. Let's find the abscissas of the points of intersection of the parabola with the $O x$ axis, for which we solve the equation $x^{2}-x=0$, from which $x_{1}=0$ and $x_{2}=1$. The angle of the curve with the $O x$ axis is the angle that the tangent, drawn at the point of intersection of the curve with the $O x$ ... | \alpha_{1}=\frac{3\pi}{4},\alpha_{2}=\frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,023 |
378. At what angle does the parabola $y=x^{2} / 2$ intersect the line $3 x-2 y-2=0$? | Solution. This question can be rephrased as: at what angle do the tangents to the curve $y=x^{2} / 2$, drawn at the points of intersection with the line $3 x-2 y-2=0$, and the line itself intersect?
We find the points of intersection of the parabola $y=x^{2} / 2$ and the line $3 x-2 y-2=0$ by solving the system of equ... | \varphi_{1}=\operatorname{arctg}(1/5),\varphi_{2}=\operatorname{arctg}(1/8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,024 |
379. Form the equation of the tangent to the parabola $y=x^{2}$ $-4 x$ at the point with abscissa $x_{0}=1$. | Solution. Let's determine the ordinate $y_{0}$ of the point of tangency by substituting the abscissa value $x_{0}=1$ into the equation of the parabola; we have $y_{0}=1-4 \cdot 1=$ $=-3$.
To find the slope of the tangent, we compute the value of the derivative at the point of tangency: $f^{\prime}(x)=2 x-4 ; k=f^{\pri... | -2x-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,025 |
380. Form the equations of the tangent and normal to the curve $y=\frac{1}{1+x^{2}}$ at the point $\left(2 ; \frac{1}{5}\right)$ | Solution. The equation of the pencil of lines passing through the point $\left(2 ; \frac{1}{5}\right)$ is $y-\frac{1}{5}=k(x-2)$. We find the slope of the tangent:
$$
k_{1}=y_{x=2}^{\prime}=\left(\frac{1}{1+x^{2}}\right)_{x-2}^{\prime}=-\left.\frac{2 x}{\left(1+x^{2}\right)^{2}}\right|_{x=2}=-\frac{4}{25}
$$
Since th... | 4x+25y-13=0125x-20y-246=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,026 |
381. Given the curve $y=-x^{2}+4$. Draw a tangent to it at the point where the abscissa is $x=-1$. | Solution. Let's find the ordinate of the point of tangency: $y_{x=-1}=\left(-x^{2}+\right.$ $+4)_{x=-1}=-(-1)^{2}+4=3$; hence, $A(-1 ; 3)$ is the point of tangency.
The equation of any line passing through point $A$ is $y-3=k(x+1)$. For the line to be a tangent, it is necessary and sufficient that $k=y_{x=-1}^{\prime}... | 2x+5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,027 |
382. At what point is the tangent to the curve $y=\frac{1}{3} x^{3}-3 x^{2}+$ $+8 x+4$ parallel to the line $2 x+2 y-5=0$? | Solution. The slope of the given line $k_{1}=-1$. The slope of the tangent $k=y^{\prime}=x^{2}-6 x+8$.
From the condition of parallelism, it follows that $k=k_{1}$. Then $x^{2}-6 x+8=-1$; $x^{2}-6 x+9=0 ; \quad(x-3)^{2}=0$. Therefore, $x=3$ is the abscissa of the point of tangency. Substituting this value of $x$ into ... | (3,10) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,028 |
383. At what angle do the curves $y=2^{x}$ and $y=$ $=\sqrt{x+1}$ intersect? | Solution. Let's find the points of intersection of the given curves; by solving the equation $2^{x}=\sqrt{x+1}$, we get $x_{1}=-0.5, x_{2}=0$.
The angle between the curves is understood as the angle between the tangents at the point of intersection of the curves. The magnitude of the angle is found using the known for... | \varphi=\operatorname{arctg}\frac{(1-\ln2)\sqrt{2}}{2+\ln2}at-0.5,\varphi=\operatorname{arctg}\frac{1-2\ln2}{2+\ln2}at0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,029 |
405. The path traveled by a material point is given by the following function of time: $s=3 t^{2}-2 t+4$. Find the speed of the point at the end of the 5th second. | Solution. We find the derivative: $\frac{d s}{d t}=6 t-2$; at $t=5$ we get $\frac{d s}{d t}=6 \cdot 5-2=28(\mathrm{m} / \mathrm{s})$. Therefore, $v=\left.\frac{d s}{d t}\right|_{t=5}=28 \mathrm{m} / \mathrm{s}$. | 28\mathrm{}/\mathrm{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,030 |
410. The height of a body thrown vertically upwards changes with time according to the law $H=200 t-4.9 t^{2}$. Find the velocity of the body at the end of the 10th second. How many seconds will the body fly upwards and what is the greatest height it will reach? | Solution. The velocity of the body is determined by the expression $v=\frac{d H}{d t}=$ $=200-9.8 t$; at $t=10$ we have $v=102($ m $/ \mathrm{s})$.
At the moment when the body reaches its maximum height, its velocity is zero. Therefore, for this moment $\frac{d H}{d t}=200-$ $-9.8 t=0$, from which $t=\frac{200}{9.8} \... | 102 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,031 |
413. A body with a mass of 8 kg moves in a straight line according to the law $s=$ $=2 t^{2}+3 t-1$. Find the kinetic energy of the body $\left(m v^{2} / 2\right)$ 3 seconds after the start of the motion. | Solution. Let's find the velocity of the body at any moment of time $t$:
$$
v=\frac{d s}{d t}=4 t+3
$$
Calculate the velocity of the body at the moment $t=3$:
$$
v_{t=3}=4 \cdot 3+3=15(\mathrm{M} / \mathrm{c})
$$
Determine the kinetic energy of the body at the moment $t=3$:
$$
\frac{m v^{2}}{2}=\frac{8 \cdot 15^{2... | 900 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,032 |
433. A point moves in a straight line according to the law $s=3 t^{2}-3 t+$ +8 . Find the velocity and acceleration of the point at the moment $t=4$. | Solution. To determine the velocity, we need to find the first derivative of the given function at $t=4$. We have
$$
v=s^{\prime}=\left(3 t^{2}-3 t+8\right)^{\prime}=6 t-3 ; v_{t=4}=6 \cdot 4-3=21(\mathrm{~m} / \mathrm{s})
$$
The acceleration is equal to the second derivative of the function at $t=4$, i.e.,
$$
a=s^{... | v_{=4}=21\,\mathrm{}/\mathrm{},\,=6\,\mathrm{}/\mathrm{}^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,033 |
434. A material point moves according to the law $s=2 t^{3}-6 t^{2}+$ $+4 t$. Find its acceleration at the end of the 3rd second. | Solution. We find $v=s^{\prime}=6 t^{2}-12 t+4, a=s^{\prime \prime}=12 t-12$, from which at $t=3$ we get $a=12 \cdot 3-12=24\left(\mathrm{m} / \mathrm{c}^{2}\right)$. | 24 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,034 |
435. At time $t$, the body is at a distance of $s=\frac{1}{4} t^{4}+4 t^{3}+16 t^{2}$ km from the starting point. Find its acceleration after 2 hours. | Solution. We find $v=s^{\prime}=t^{3}+12 t^{2}+32 t, a=v^{\prime}=s^{\prime \prime}=3 t^{2}+$ $+24 t+32$. For $t=2$ we have $a=3 \cdot 4+24 \cdot 2+32=92$ (km $\left./ \mathrm{u}^{2}\right)$. | 92 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,035 |
440. A point moves along the x-axis according to the law $x=2 t^{3}-4 t^{2}+2 t+3$, where $t$ is time in seconds measured from $t=0$, and $x$ is the distance of the moving point from the origin in meters. It is required: a) to determine the law of variation of velocity and acceleration with time $t$; b) to find the ini... | Solution. a) To determine the velocity of the motion, we find the derivative of the path with respect to time:
$$
v=\frac{d x}{d t}=6 t^{2}-8 t+2=2\left(3 t^{2}-4 t+1\right)
$$
and for determining the acceleration of the motion - the derivative of the velocity with respect to time:
$$
a=\frac{d v}{d t}=2(5 t-4)=4(3 ... | 3\frac{8}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,036 |
448. The sides $a$ and $b$ of a rectangle change according to the laws $a=(2 t+1)$ cm, $b=(3 t+2)$ cm. At what rate is its area $S$ changing at the moment $t=4$ s? | Solution. We find $S=a b=(2 t+1)(3 t+2)=6 t^{2}+7 t+2, \quad v=$ $=S_{t}^{\prime}=\left(6 t^{2}+7 t+2\right)^{\prime}=12 t+7$. For $t=4$ we get $v_{t=4}=12 \cdot 4+7=$ $=55(\mathrm{~cm} / \mathrm{s})$. | 55 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,037 |
455. The amount of electricity flowing through a conductor, starting from the moment of time $t=0$, is given by the formula $Q=$ $=3 t^{2}-3 t+4$. Find the current strength at the end of the 6th second. | Solution. The current is the derivative of the quantity of electricity with respect to time: therefore, we need to find the derivative of the function $Q=3 t^{2}-3 t+4$ and calculate its value at $t=6 \mathrm{c}$. We have $I=Q^{\prime}=$ $=6 t-3$, from which at $t=6$ we get $I=33$ (A). | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,039 |
460. The amount of heat $Q$, received by a certain substance when heated from 0 to $T$, is determined by the formula $Q=$ $=0.1054 t+0.000002 t^{2}$ ( $Q$ - in joules, $t$ - in kelvins). Find the heat capacity of this substance at $100 \mathrm{~K}$. | Solution. We find the heat capacity:
$$
C=Q_{T}^{\prime}=0.1054+0.000004 \dot{T}
$$
At $T=100 \mathrm{~K}$ we get
$$
C_{T=100}=0.1054+0.000004 \cdot 100=0.1058(\text{J/K). }
$$ | 0.1058 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,040 |
461. The law of temperature change $T$ of a body depending on time $t$ is given by the equation $T=0.2 t^{2}$. At what rate is this body heating at the moment of time $10 \mathrm{c}$? | Solution. The heating rate of a body is the derivative of temperature $T$ with respect to time $t$:
$$
\frac{d T}{d t}=\left(0.2 t^{2}\right)^{\prime}=0.4 t
$$
Determine the heating rate of the body at $t=10$:
$$
\left(\frac{d T}{d t}\right)_{t=10}=0.4 \cdot 10=4(\text{ deg } / \mathrm{s})
$$ | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,041 |
463. Find the increment and differential of the function $y=x^{2}$ at the point $x=2$ for $\Delta x=0.1$. | Solution. $\quad \Delta y=(x+\Delta x)^{2}-x^{2}=x^{2}+2 x \Delta x+(\Delta x)^{2}-x^{2}=2 x \Delta x+(\Delta x)^{2}$;
$$
\begin{aligned}
& \left.\Delta y\right|_{\Delta x=0.1}=2 \cdot 2 \cdot 0.1+(0.1)^{2}=0.4+0.01=0.41 \\
& d y=y^{\prime} \cdot \Delta x=\left(x^{2}\right)^{\prime} \cdot \Delta x=2 x \cdot \Delta x ;... | 0.410.4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,042 |
467. Find the differentials of the functions:
a) $y=\sqrt{x^{3}-3}$
b) $y=\frac{x-1}{x+2}$
c) $y=x(x+1)$. | Solution. a) $d y=\frac{3 x^{2}}{2 \sqrt{x^{3}-3}} d x$;
b) $d y=\frac{(x-1)^{\prime}(x+2)-(x+2)^{\prime}(x-1)}{(x+2)^{2}} d x=\frac{(x+2)-(x-1)}{(x+2)^{2}} d x=$ $=\frac{3}{(x+2)^{2}} d x=\frac{3 d x}{(x+2)^{2}}$
c) $d y=\left(x^{\prime}(x+1)+(x+1)^{\prime} x\right) d x=(x+1+x) d x=(2 x+1) d x$.
468-477. Find the d... | )\frac{3x^{2}}{2\sqrt{x^{3}-3}};\quadb)\frac{3}{(x+2)^{2}};\quad)(2x+1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,043 |
490. Using the concept of the differential of a function, approximately calculate the change in the function $y=x^{3}-7 x^{2}+80$ when the argument $x$ changes from 5 to 5.01. | Solution. We find
$$
\Delta y \approx d y=y^{\prime} \Delta x=\left(3 x^{2}-14 x\right) \Delta x
$$
For $x=5, \Delta x=5.01-5=0.01$, we get
 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,044 | |
496. Find approximately the increment of the function $y=3 x^{2}+2$ at $x=2$ and $\Delta x=0.001$. Determine the absolute and relative errors of the calculation. | Solution. Since the increment of the argument is a small quantity, the increment of the function can be replaced by its differential:
$$
\left.\Delta y \approx d y\right|_{\substack{x=2 \\ d x=0.001}}=\left.6 x d x\right|_{\substack{x=2, d x=0.001}}=6 \cdot 2 \cdot 0.001=0.012
$$
Let's find the error obtained by repl... | 0.012003,0.000003,0.025 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,045 |
497. Find the absolute and relative errors when replacing the increment of the function $y=x^{3}+2 x$ with its differential at the point $x=2$ for $\Delta x=0.1$. | Solution. $\Delta y=\left((x+\Delta x)^{3}+2(x+\Delta x)\right)-\left(x^{3}+2 x\right)=3 x^{2} \Delta x+$ $+3 x(\Delta x)^{2}+(\Delta x)^{3}+2 \Delta x$
$$
\begin{aligned}
& \left.\Delta y\right|_{\substack{x=2,0,1 \\
\Delta x=0,1}}=3 \cdot 2^{2} \cdot 0.1+3 \cdot 2 \cdot 0.1^{2}+0.1^{3}+2 \cdot 0.1=1.461 ; \\
& d y=y... | 0.061,4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,046 |
501. Using the differential, calculate the increment of the function $y=x \sqrt{x^{2}+5}$ with an accuracy of 0.01 when $x=2$ and $\Delta x=0.2$. | Solution. We find the differential of the given function:
$$
d y=y^{\prime} d x=\left(\sqrt{x^{2}+5}+\frac{x^{2}}{\sqrt{x^{2}+5}}\right) d x
$$
For $x=2$ and $\Delta x=0.2$, we get
$$
\left.\Delta y \approx d y\right|_{\substack{x=2 . \\ d x=0,2}}=\left(\sqrt{4+5}+\frac{4}{\sqrt{4+5}}\right) \cdot 0.2 \approx 0.866 ... | 0.87 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,047 |
503. By direct measurement, it was found that the diameter of a circle is 6.4 cm, with the maximum error not exceeding $0.05 \mathrm{~cm}$. Find approximately the maximum error in the estimation of the area, calculated by the formula $S=\frac{1}{4} \pi x^{2}$ ( $x$ - diameter). | Solution. Obviously, the exact value of the maximum error is $\Delta S$, where $x$ varies from 6.4 to 6.45. The approximate value of the maximum error is the corresponding differential $d S$. We find
$$
d S=S^{\prime}(x) d x=\left.\frac{1}{2} \pi x d x\right|_{\substack{x=6.4 \\ d x=0.05}}=\frac{1}{2} \pi \cdot 6.4 \c... | 0.5024(\mathrm{~}^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,048 |
510. Find the approximate value of the function $y=\sqrt{3 x^{2}+1}$ at $x=1.02$. | Solution. We will use formula (6), i.e., $f(x+\Delta x) \approx$
$\approx f(x)+d y$. In this case, we should take $x=1$. Then $\Delta x=$ $=1.02-1=0.02$. The value of the function at $x=1$ is easily determined: $f(1)=\sqrt{3 \cdot 1^{2}+1}=2$. Next, we find
$$
d y=\left(\sqrt{3 x^{2}+1}\right)^{\prime} d x=\frac{6 x d... | 2.03 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,049 |
511. Calculate $\operatorname{tg} 46^{\circ}$, based on the value of the function $y=\operatorname{tg} x$ at $x=45^{\circ}$ and replacing its increment with a differential. | Solution. Here $x=45^{\circ}, \Delta x=46^{\circ}-45^{\circ}=1^{\circ}$. To use formula (6), we need to express the angles in radians: $x=\pi / 4, \Delta x=$ $=0.0175$.
We calculate the differential of the function at $x=\pi / 4$ and $d x=\Delta x=0.0175$:
$$
d y=\left.\frac{1}{\cos ^{2} x} d x\right|_{\substack{x=\p... | 1.0350 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,050 |
512. Find the approximate value of $\sqrt[5]{3} 1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The note is not part of the translation but is provided to clarify the task. The actual translation is as follows:
512.... | Solution. Consider the function $y=\sqrt[5]{x}$ and find its differential: $d y=\frac{1}{5 \sqrt[5]{x^{4}}} d x$. Calculate the increment of the function $\Delta y$ when $x$ changes from 32 to 31, i.e., when $\Delta x=-1$:
$$
\Delta y \approx d y=\left.\frac{1}{5 \sqrt[5]{x^{1}}} d x\right|_{\substack{x=32 \\ d x=-1}}... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,051 | |
513. Find the approximate value of the function $y=\sqrt[3]{x^{2}+2 x}$ at $x=2.004$, based on its exact value at $x_{0}=2$ and replacing $\Delta y$ with $d y$. | Solution. Here the initial value $x_{0}=2$, the incremented value $x=x_{0}+\Delta x=2.004, \Delta x=2.004-2=0.004$.
Let's find the initial value of the function: $\left.y\right|_{x=2}=\sqrt[3]{2^{2}+4}=\sqrt[3]{8}=2$.
Let's find the differential:
$$
d y=\left(\sqrt[3]{x^{2}+2 x}\right)^{\prime} d x=\left(\left(x^{2}... | 2.002 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,052 |
514. Calculate approximately $3.002^{4}$. | Solution. Consider the function $f(x)=x^{4}$. Here $x_{1}=3.002, x_{0}=3$, $\Delta x=0.002$ and, therefore, $f\left(x_{1}\right)=f(3.002)=3.002^{4}$. We find $f\left(x_{0}\right)=$ $=f(3)=3^{4}=81, f^{\prime}\left(x_{0}\right) \Delta x=4 \cdot 3^{3} \cdot 0.002=4 \cdot 27 \cdot 0.002=0.216$. Thus, $3.002^{4} \approx$ $... | 81.216 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,053 |
515. Calculate approximately $1.998^{5}$. | Solution. Consider the function $f(x)=x^{5}$. Here $x_{1}=1.998$, $x_{0}=2, \quad \Delta x=-0.002, f\left(x_{1}\right)=f(1.998)=1.998^{5}$. Next, we find $f\left(x_{0}\right)=$ $=f(2)=2^{5}=32, f^{\prime}\left(x_{0}\right) \Delta x=5 \cdot 2^{4}(-0.002)=5 \cdot 16(-0.002)=-0.16$.
As a result, we get $1.998^{5} \approx... | 31.84 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,054 |
516. Find approximately $\sin 31^{\circ}$. | Solution. We will find an approximate value of the function $\boldsymbol{y}=$ $=\sin x$ at $x=31^{\circ}$, based on its exact value at $x=30^{\circ}$. We have $\sin 30^{\circ}=0.5, d x \approx \Delta x=31^{\circ}-30^{\circ}=1^{\circ}$ (or in radians $d x=0.0175$).
Let's find $d y=\cos x d x$. Further, at $x=30^{\circ}... | 0.515 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,055 |
536. Show that the function $y=2 x^{3}+3 x^{2}-12 x+1$ is decreasing in the interval $(-2,1)$. | Solution. It is sufficient to verify that the derivative of the function is negative for $-2<x<1$. We find
$$
y^{\prime}=6 x^{2}+6 x-12=6(x+2)(x-1)
$$
The factor $x+2$ is positive on the interval $(-2,1)$, while the factor $x-1$ is negative. Therefore, the derivative is negative at all points in the specified interva... | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,056 |
537. Show that the function $y=\tan x$ is monotonically increasing in the interval $(-\pi / 2, \pi / 2)$. | Solution. We find the derivative $(\operatorname{tg} x)^{\prime}=\frac{1}{\cos ^{2} x}$. In the given interval, $\cos x$ varies from 0 to $1 ; ;$ therefore, $(\operatorname{tg} x)^{\prime}=\frac{1}{\cos ^{2} x}>0$. Consequently, the given function is increasing. | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,057 |
538. Investigate the behavior of the function $f(x)=3+\sqrt{x}$ at the point $x=4$. | Solution. Let's find the derivative; $f^{\prime}(x)=\frac{1}{2 \sqrt{x}}$ Calculate the value of the derivative at $x=4: f^{\prime}(4)=\frac{1}{2 \sqrt{4}}=\frac{1}{4}>0$. From this, we conclude that the given function is increasing at the point $x=4$. | f^{\}(4)=\frac{1}{4}>0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,058 |
542. Show that the function $y=\frac{x^{2}-1}{x}$ is increasing in any interval that does not contain the point $x=0$.
We have established that the intervals of increase or decrease of a function coincide with the intervals in which the derivative of this function maintains its sign. Therefore, a transition from incre... | Solution. $1^{0}$. Find the derivative of the given function: $y^{\prime}=2 x-4$.
Fig. 115
=\frac{\sin x}{x}$ on the interval $\left(0, \frac{\pi}{2}\right)$. | solution. We have
$$
f^{\prime}(x)=\left(\frac{\sin x}{x}\right)=\frac{x \cos x-\sin x}{x^{2}}=\frac{x-\operatorname{tg} x}{x^{2} \cos x}
$$
If $0<x<\pi / 2$, then $\cos x>0$ and $x-\operatorname{tg} x<0$, so $f^{\prime}(x)<0$. It follows that $f(x)$ is decreasing on $(0, \pi / 2)$. | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,061 |
564. $y=x^{2}+2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
564. $y=x^{2}+2$. | Solution. $1^{0}$. Find the derivative: $y^{\prime}=\left(x^{2}+2\right)^{\prime}=2 x$.
$2^{0}$. Set it to zero; $2 x=0$, from which $x=0$ - the critical point.
$3^{\circ}$. Determine the sign of the derivative at the value $x0$, for example at $x=1$: $y_{x=1}^{\prime}=2 \cdot 1=2$. Since the derivative changes sign ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,064 |
565. $y=\frac{1}{3} x^{3}-2 x^{2}+3 x+1$. | Solution. $1^{0}$. Find the derivative $y^{\prime}=x^{2}-4 x+3$.
$2^{0}$. Set it to zero and solve the equation $x^{2}-4 x+3=0$. Its roots $x_{1}=1, x_{2}=3$ are critical points.
$3^{0}$. The derivative can be represented as a product of factors: $y^{\prime}=(x-1)(x-3)$. Investigate the critical point $x_{1}=1$, dete... | 0.y|_{1}=\frac{7}{3},.y|_{3}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,065 |
566. $f(x)=x^{3}+3 x^{2}+9 x-6$.
566. $f(x)=x^{3}+3 x^{2}+9 x-6$. | Solution. Find $f^{\prime}(x)=3 x^{2}+6 x+9=3\left(x^{2}+2 x+3\right)$. Setting
Fig. 122 the derivative to zero, we see that the equation $x^{2}+2 x+3=0$ has no real roots, which means that ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,066 |
567. $y=(x-5) e^{x}$. | Solution. $1^{0} . y^{\prime}=(x-5)^{\prime} e^{x}+$ $+\left(e^{x}\right)^{\prime}(x-5)=e^{x}+e^{x}(x-5)=e^{x}(x-4)$. $2^{0} . e^{x}(x-4)=0 ; e^{x} \neq 0 ; x-4=0 ; x=4$.
$3^{\prime \prime}$. In the interval $(-\infty, 4)$ the derivative is negative, and in the interval $(4, \infty)$ it is positive: Therefore, at $x=4... | -e^{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,067 |
568. $y=1-\sqrt[5]{(x-2)^{4}}$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
568. $y=1-\sqrt[5]{(x-2)^{4}}$ | Solution. $1^{0} . y^{\prime}=-\frac{4}{5}(x-2)^{-1 / 5}=-\frac{4}{5 \sqrt[5]{x-2}}$.
$2^{0}$. The derivative does not become zero for any values of $x$ and does not exist only at $x=2$. This is the critical point.
$3^{0}$. In the interval $(-\infty, 2)$, the derivative is positive, and in the interval $(2, \infty)$,... | y_{\max}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,068 |
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