text
stringlengths
2
2.33k
source
stringclasses
826 values
We can use a formula to find the derivative ofy=lnx,y=lnx,and the relationshiplogbx=lnxlnblogbx=lnxlnballows us to extend our differentiation formulas to include logarithms with arbitrary bases.
https://openstax.org/books/calculus-volume-1/pages/3-key-concepts
Logarithmic differentiation allows us to differentiate functions of the formy=g(x)f(x)y=g(x)f(x)or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.
https://openstax.org/books/calculus-volume-1/pages/3-key-concepts
Q = f ( x ) − f ( a ) x − a Q = f ( x ) − f ( a ) x − a
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
Q = f ( a + h ) − f ( a ) a + h − a = f ( a + h ) − f ( a ) h Q = f ( a + h ) − f ( a ) a + h − a = f ( a + h ) − f ( a ) h
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
m tan = lim x → a f ( x ) − f ( a ) x − a m tan = lim x → a f ( x ) − f ( a ) x − a m tan = lim h → 0 f ( a + h ) − f ( a ) h m tan = lim h → 0 f ( a + h ) − f ( a ) h
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
f ′ ( a ) = lim x → a f ( x ) − f ( a ) x − a f ′ ( a ) = lim x → a f ( x ) − f ( a ) x − a f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
v a ve = s ( t ) − s ( a ) t − a v a ve = s ( t ) − s ( a ) t − a
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
v ( a ) = s ′ ( a ) = lim t → a s ( t ) − s ( a ) t − a v ( a ) = s ′ ( a ) = lim t → a s ( t ) − s ( a ) t − a
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( sin x ) = cos x d d x ( sin x ) = cos x
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( cos x ) = − sin x d d x ( cos x ) = − sin x
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( tan x ) = sec 2 x d d x ( tan x ) = sec 2 x
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( cot x ) = − csc 2 x d d x ( cot x ) = − csc 2 x
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( sec x ) = sec x tan x d d x ( sec x ) = sec x tan x
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( csc x ) = − csc x cot x d d x ( csc x ) = − csc x cot x
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
h ′ ( x ) = f ′ ( g ( x ) ) g ′ ( x ) h ′ ( x ) = f ′ ( g ( x ) ) g ′ ( x )
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
h ′ ( x ) = n ( g ( x ) ) n − 1 g ′ ( x ) h ′ ( x ) = n ( g ( x ) ) n − 1 g ′ ( x )
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
( f −1 ) ′ ( x ) = 1 f ′ ( f −1 ( x ) ) ( f −1 ) ′ ( x ) = 1 f ′ ( f −1 ( x ) ) whenever f ′ ( f −1 ( x ) ) ≠0 f ′ ( f −1 ( x ) ) ≠0 and f ( x ) f ( x ) is differentiable.
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( x m / n ) = m n x ( m / n ) − 1 . d d x ( x m / n ) = m n x ( m / n ) − 1 .
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x sin −1 x = 1 1 − ( x ) 2 d d x sin −1 x = 1 1 − ( x ) 2
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x cos −1 x = −1 1 − ( x ) 2 d d x cos −1 x = −1 1 − ( x ) 2
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x tan −1 x = 1 1 + ( x ) 2 d d x tan −1 x = 1 1 + ( x ) 2
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x cot −1 x = −1 1 + ( x ) 2 d d x cot −1 x = −1 1 + ( x ) 2
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x sec −1 x = 1 | x | ( x ) 2 − 1 d d x sec −1 x = 1 | x | ( x ) 2 − 1
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x csc −1 x = −1 | x | ( x ) 2 − 1 d d x csc −1 x = −1 | x | ( x ) 2 − 1
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( e g ( x ) ) = e g ( x ) g ′ ( x ) d d x ( e g ( x ) ) = e g ( x ) g ′ ( x )
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( ln g ( x ) ) = 1 g ( x ) g ′ ( x ) d d x ( ln g ( x ) ) = 1 g ( x ) g ′ ( x )
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( b g ( x ) ) = b g ( x ) g ′ ( x ) ln b d d x ( b g ( x ) ) = b g ( x ) g ′ ( x ) ln b
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
d d x ( log b g ( x ) ) = g ′ ( x ) g ( x ) ln b d d x ( log b g ( x ) ) = g ′ ( x ) g ( x ) ln b
https://openstax.org/books/calculus-volume-1/pages/3-key-equations
acceleration : is the rate of change of the velocity, that is, the derivative of velocity
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
amount of change : the amount of a functionf(x)f(x)over an interval[x,x+h][x,x+h]isf(x+h)−f(x)f(x+h)−f(x)
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
average rate of change : is a functionf(x)f(x)over an interval[x,x+h][x,x+h]isf(a+h)−f(a)hf(a+h)−f(a)h
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
chain rule : the chain rule defines the derivative of a composite function as the derivative of the outer function evaluated at the inner function times the derivative of the inner function
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
constant multiple rule : the derivative of a constantcmultiplied by a functionfis the same as the constant multiplied by the derivative:ddx(cf(x))=cf′(x)ddx(cf(x))=cf′(x)
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
constant rule : the derivative of a constant function is zero:ddx(c)=0,ddx(c)=0,wherecis a constant
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
derivative : the slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, is the derivative
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
derivative function : gives the derivative of a function at each point in the domain of the original function for which the derivative is defined
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
difference quotient : of a functionf(x)f(x)ataais given byf(a+h)−f(a)horf(x)−f(a)x−af(a+h)−f(a)horf(x)−f(a)x−a
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
difference rule : the derivative of the difference of a functionfand a functiongis the same as the difference of the derivative offand the derivative ofg:ddx(f(x)−g(x))=f′(x)−g′(x)ddx(f(x)−g(x))=f′(x)−g′(x)
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
differentiable ata : a function for whichf′(a)f′(a)exists is differentiable ataa
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
differentiable function : a function for whichf′(x)f′(x)exists is a differentiable function
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
differentiable onS : a function for whichf′(x)f′(x)exists for eachxxin the open setSSis differentiable onSS
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
differentiation : the process of taking a derivative
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
higher-order derivative : a derivative of a derivative, from the second derivative to thenth derivative, is called a higher-order derivative
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
implicit differentiation : is a technique for computingdydxdydxfor a function defined by an equation, accomplished by differentiating both sides of the equation (remembering to treat the variableyyas a function) and solving fordydxdydx
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
instantaneous rate of change : the rate of change of a function at any point along the functiona,a,also calledf′(a),f′(a),or the derivative of the function ataa
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
logarithmic differentiation : is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
marginal cost : is the derivative of the cost function, or the approximate cost of producing one more item
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
marginal profit : is the derivative of the profit function, or the approximate profit obtained by producing and selling one more item
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
marginal revenue : is the derivative of the revenue function, or the approximate revenue obtained by selling one more item
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
population growth rate : is the derivative of the population with respect to time
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
power rule : the derivative of a power function is a function in which the power onxxbecomes the coefficient of the term and the power onxxin the derivative decreases by 1: Ifnnis an integer, thenddxxn=nxn−1ddxxn=nxn−1
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
product rule : the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function:ddx(f(x)g(x))=f′(x)g(x)+g′(x)f(x)ddx(f(x)g(x))=f′(x)g(x)+g′(x)f(x)
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
quotient rule : the derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function:ddx(f(x)g(x))=f′(x)g(x)−g′(x)f(x)(g(x))2ddx(f(x)g(x))=f′(x)g(x)...
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
speed : is the absolute value of velocity, that is,|v(t)||v(t)|is the speed of an object at timettwhose velocity is given byv(t)v(t)
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
sum rule : the derivative of the sum of a functionfand a functiongis the same as the sum of the derivative offand the derivative ofg:ddx(f(x)+g(x))=f′(x)+g′(x)ddx(f(x)+g(x))=f′(x)+g′(x)
https://openstax.org/books/calculus-volume-1/pages/3-key-terms
To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related quantities that are changing with respect to time.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
In terms of the quantities, state the information given and the rate to be found.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Find an equation relating the quantities.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Use differentiation, applying the chain rule as necessary, to find an equation that relates the rates.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
A differentiable functiony=f(x)y=f(x)can be approximated ataaby the linear functionL(x)=f(a)+f′(a)(x−a).L(x)=f(a)+f′(a)(x−a).
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
For a functiony=f(x),y=f(x),ifxxchanges fromaatoa+dx,a+dx,thendy=f′(x)dxdy=f′(x)dxis an approximation for the change iny.y.The actual change inyyisΔy=f(a+dx)−f(a).Δy=f(a+dx)−f(a).
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
A measurement errordxdxcan lead to an error in a calculated quantityf(x).f(x).The error in the calculated quantity is known as thepropagated error. The propagated error can be estimated bydy≈f′(x)dx.dy≈f′(x)dx.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
To estimate the relative error of a particular quantityq,q,we estimateΔqq.Δqq.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
A function may have both an absolute maximum and an absolute minimum, have just one absolute extremum, or have no absolute maximum or absolute minimum.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
If a function has a local extremum, the point at which it occurs must be a critical point. However, a function need not have a local extremum at a critical point.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
A continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. Each extremum occurs at a critical point or an endpoint.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Ifffis continuous over[a,b][a,b]and differentiable over(a,b)(a,b)andf(a)=f(b),f(a)=f(b),then there exists a pointc∈(a,b)c∈(a,b)such thatf′(c)=0.f′(c)=0.This is Rolle’s theorem.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Ifffis continuous over[a,b][a,b]and differentiable over(a,b),(a,b),then there exists a pointc∈(a,b)c∈(a,b)such thatf′(c)=f(b)−f(a)b−a.f′(c)=f(b)−f(a)b−a.This is the Mean Value Theorem.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Iff′(x)=0f′(x)=0over an intervalI,I,thenffis constant overI.I.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
If two differentiable functionsffandggsatisfyf′(x)=g′(x)f′(x)=g′(x)overI,I,thenf(x)=g(x)+Cf(x)=g(x)+Cfor some constantC.C.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Iff′(x)>0f′(x)>0over an intervalI,I,thenffis increasing overI.I.Iff′(x)<0f′(x)<0overI,I,thenffis decreasing overI.I.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Ifccis a critical point offfandf′(x)>0f′(x)>0forx<cx<candf′(x)<0f′(x)<0forx>c,x>c,thenffhas a local maximum atc.c.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Ifccis a critical point offfandf′(x)<0f′(x)<0forx<cx<candf′(x)>0f′(x)>0forx>c,x>c,thenffhas a local minimum atc.c.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Iff″(x)>0f″(x)>0over an intervalI,I,thenffis concave up overI.I.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Iff″(x)<0f″(x)<0over an intervalI,I,thenffis concave down overI.I.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Iff′(c)=0f′(c)=0andf″(c)>0,f″(c)>0,thenffhas a local minimum atc.c.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Iff′(c)=0f′(c)=0andf″(c)<0,f″(c)<0,thenffhas a local maximum atc.c.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Iff′(c)=0f′(c)=0andf″(c)=0,f″(c)=0,then evaluatef′(x)f′(x)at a test pointxxto the left ofccand a test pointxxto the right ofc,c,to determine whetherffhas a local extremum atc.c.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
The limit off(x)f(x)isLLasx→∞x→∞(or asx→−∞)x→−∞)if the valuesf(x)f(x)become arbitrarily close toLLasxxbecomes sufficiently large.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
The limit off(x)f(x)is∞∞asx→∞x→∞iff(x)f(x)becomes arbitrarily large asxxbecomes sufficiently large. The limit off(x)f(x)is−∞−∞asx→∞x→∞iff(x)<0f(x)<0and|f(x)||f(x)|becomes arbitrarily large asxxbecomes sufficiently large. We can define the limit off(x)f(x)asxxapproaches−∞−∞similarly.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
For a polynomial functionp(x)=anxn+an−1xn−1+…+a1x+a0,p(x)=anxn+an−1xn−1+…+a1x+a0,whereanâ‰0,anâ‰0,the end behavior is determined by the leading termanxn.anxn.Ifnâ‰0,nâ‰0,p(x)p(x)approaches∞∞or−∞−∞at each end.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
For a rational functionf(x)=p(x)q(x),f(x)=p(x)q(x),the end behavior is determined by the relationship between the degree ofppand the degree ofq.q.If the degree ofppis less than the degree ofq,q,the liney=0y=0is a horizontal asymptote forf.f.If the degree ofppis equal to the degree ofq,q,then the liney=anbny=anbnis a ho...
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
To solve an optimization problem, begin by drawing a picture and introducing variables.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Find an equation relating the variables.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Find a function of one variable to describe the quantity that is to be minimized or maximized.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Look for critical points to locate local extrema.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form0000or∞/∞∞/∞arises.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form0000or∞/∞.∞/∞.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
The exponential functionexexgrows faster than any power functionxp,xp,p>0.p>0.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
The logarithmic functionlnxlnxgrows more slowly than any power functionxp,xp,p>0.p>0.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Newton’s method approximates roots off(x)=0f(x)=0by starting with an initial approximationx0,x0,then uses tangent lines to the graph offfto create a sequence of approximationsx1,x2,x3,….x1,x2,x3,….
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Typically, Newton’s method is an efficient method for finding a particular root. In certain cases, Newton’s method fails to work because the list of numbersx0,x1,x2,…x0,x1,x2,…does not approach a finite value or it approaches a value other than the root sought.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Any process in which a list of numbersx0,x1,x2,…x0,x1,x2,…is generated by defining an initial numberx0x0and defining the subsequent numbers by the equationxn=F(xn−1)xn=F(xn−1)for some functionFFis an iterative process. Newton’s method is an example of an iterative process, where the functionF(x)=x−[f(x)f′...
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
IfFFis an antiderivative off,f,then every antiderivative offfis of the formF(x)+CF(x)+Cfor some constantC.C.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
Solving the initial-value problemdydx=f(x),y(x0)=y0dydx=f(x),y(x0)=y0requires us first to find the set of antiderivatives offfand then to look for the particular antiderivative that also satisfies the initial condition.
https://openstax.org/books/calculus-volume-1/pages/4-key-concepts
L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L ( x ) = f ( a ) + f ′ ( a ) ( x − a )
https://openstax.org/books/calculus-volume-1/pages/4-key-equations
d y = f ′ ( x ) d x . d y = f ′ ( x ) d x .
https://openstax.org/books/calculus-volume-1/pages/4-key-equations
absolute extremum : ifffhas an absolute maximum or absolute minimum atc,c,we sayffhas an absolute extremum atcc
https://openstax.org/books/calculus-volume-1/pages/4-key-terms