Selecting Procedures for Calculating Derivatives (College Board AP® Calculus BC): Study Guide

Selecting procedures for calculating derivatives

• You should be familiar with all the different methods for differentiating functions

• This way you can choose the most appropriate method to use for an exam question

• You should also know that the derivative of $f\left(x\right)$ is defined as

  • $f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  • You generally only need to use this if asked to do so

Estimating derivatives

• You can estimate a derivative at a point using a table or graph

  • Remember that the derivative at a point is equal to the slope of the tangent at that point

  • To approximate the slope of the tangent to the graph of $f\left(x\right)$ at $x=a$:

    • Find the slope of line segments joining nearby coordinates that lie on the graph

  • This is useful when you only have a graph or table

Finding the derivative as an expression

Basic differentiation

• If you know the function or the equation of the graph, you can differentiate it using several methods

• Powers of $x$ are differentiated according to the following formula:

  • If$f\left(x\right)=x^n$ then$f^{\text{'}}\left(x\right)=n{x}^{n-1}$ 

• When differentiating sums or differences of powers of $x$,

  • the derivative is simply the sum (or difference) of the derivatives of the terms

• Constant multiples of powers of $x$ are differentiated according to the following formula:

  • If$f\left(x\right)=a{x}^n$ then$f^{\text{'}}\left(x\right)=an{x}^{n-1}$ 

• Remember the two special cases

  • If $f\left(x\right)=ax$ then $f^{\text{'}}\left(x\right)=a$

  • If $g\left(x\right)=a$ then $g^{\text{'}}\left(x\right)=0$

• You may need to expand brackets or simplify expressions before differentiating

  • This may involve using laws of exponents, e.g. rewriting $\sqrt{x}$ as $x^{\frac{1}{2}}$

Differentiating exponentials and logarithms

• The table below summarises the key results for derivatives of exponentials and logarithms

Differentiating trigonometric functions

• The table below summarises the key results for derivatives of

  • trigonometric functions,

  • reciprocal trigonometric functions,

  • and inverse trigonometric functions

• The most important results to remember are for sin and cos

• The results for tan, csc, sec, and cot can all be derived using the quotient rule and trigonometric identities

• The results for the inverse trigonometric functions can be derived using either the inverse function theorem or implicit differentiation

Product rule and quotient rule

• The product rule is used when two functions are multiplied together, it states that

  • If $f\left(x\right)=g\left(x\right)·h\left(x\right)$,

  • then $f^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)·h\left(x\right)+g\left(x\right)·h^{\text{'}}\left(x\right)$

• It can also be written as,

  • if $y=u·v$,

  • then $\frac{dy}{dx}=\frac{du}{dx}·v+u·\frac{dv}{dx}$

  • Or in a more concise form: $y^{\text{'}}=u^{\text{'}}v+u{v}^{\text{'}}$

• The quotient rule is used when one function is divided by another, it states that

  • If $f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}$,

  • then $f^{\text{'}}\left(x\right)=\frac{g^{\text{'}}\left(x\right)·h\left(x\right)-g\left(x\right)·h^{\text{'}}\left(x\right)}{{\left(h\left(x\right)\right)}^2}$

• It can also be written as,

  • If $y=\frac{u}{v}$,

  • then $\frac{dy}{dx}=\frac{\frac{du}{dx}·v-u·\frac{dv}{dx}}{v^2}$

  • Or in a more concise form: $y^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

The chain rule

• The chain rule is used for composite functions (a function within a function); it states that

  • If $y=f\left(u\right)$ and $u=g\left(x\right)$ (i.e. $y$ is a function of $u$, and $u$ is a function of $x$),

  • then $\frac{dy}{dx}=\frac{dy}{du}·\frac{du}{dx}$

• Or in function notation, if $h\left(x\right)=f\left(g\left(x\right)\right)$

  • $h^{\text{'}}\left(x\right)=f^{\text{'}}\left(g\left(x\right)\right)·g^{\text{'}}\left(x\right)$

The inverse function theorem

• The inverse function theorem can be used to find the derivative of the inverse of a function, it states that

  • For a function $f$, the derivative of its inverse will be given by:

  • ${\left(f^{-1}\right)}^{\text{'}}\left(a\right)=\frac{1}{f^{\text{'}}\left(f^{-1}\left(a\right)\right)}$

• You may also see this written as:

  • $g^{\text{'}}\left(a\right)=\frac{1}{f^{\text{'}}\left(g\left(a\right)\right)}$

  • Where $g\left(a\right)=f^{-1}\left(a\right)$

• Or if $y=f^{-1}\left(x\right)$ so that $x=f\left(y\right)$,

  • then the inverse function theorem can be written as

  • $\frac{dy}{dx}=\frac{1}{\left({\displaystyle \frac{dx}{dy}}\right)}$

Implicit differentiation

• Implicit differentiation is used for functions written implicitly

  • E.g. $3x^2-7x{y}^2=3$ or $x^2+y^2=25$

• Every term in the equation is differentiated

• For terms that are only in terms of $x$, this is straightforward

• For terms that involve $y$, we apply the chain rule

  • $\frac{d}{dx}f\left(y\right)=f^{\text{'}}\left(y\right)·y^{\text{'}}=f^{\text{'}}\left(y\right)·\frac{dy}{dx}$

• In short, this means:

  • Differentiate the function that is in terms of $y$, with respect to $\mathit{y}$,

  • and then multiply it by the term $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$

• Once each term has been differentiated with respect to $x$, rearrange to make $\frac{dy}{dx}$ the subject