The Chain Rule (College Board AP® Calculus BC): Revision Note

The proof of the formula for the chain rule is beyond the scope of this course.

Be careful if you try to do the steps in your head. A common error that is frequently reported by chief readers is for students to forget to differentiate the inner functions.

For example, some students incorrectly differentiate $\sin \left(x^2\right)$ as $\cos \left(x^2\right)$. Here they have forgotten to multiply by the derivative of $x^2$.

The functions $f$ and $g$ are differentiable. The table shown gives values of the functions and their first derivatives at $x=2$.

Let $h$ be the function defined by $h\left(x\right)=f\left(g\left(x\right)\right)$. Find $h\text{'}\left(2\right)$. Show the work that leads to your answer.

Answer:

Apply the chain rule to the function $h$

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

Substitute $x=2$

$h\text{'}\left(2\right)=f\text{'}\left(g\left(2\right)\right)·g\text{'}\left(2\right)$

Use the values in the table to substitute $x=2$ into $g\left(x\right)$ and $g\text{'}\left(x\right)$

$\begin{array}{rcl}h\text{'}\left(2\right)& =& f\text{'}\left(5\right)·4\end{array}$

Use the values in the table to substitute $x=5$ into $f\text{'}\left(x\right)$

$\begin{array}{rcl}h\text{'}\left(2\right)& =& \left(-3\right)·4\end{array}$

$h\text{'}\left(2\right)=-12$

Find the derivatives of the following functions.

(a) $f\left(x\right)=\sin \left(3x^4+5x\right)$

(b) $g\left(x\right)=e^{\tan x}$

Answer:

(a)

Substitute the inside function for u

$u=3x^4+5x$

$y=\sin u$

Differentiate $u$ with respect to $x$ and differentiate $y$ with respect to $u$

$\frac{du}{dx}=12x^3+5$

$\frac{dy}{du}=\cos u$

Apply the chain rule which states $\frac{dy}{dx}=\frac{dy}{du}·\frac{du}{dx}$

$\frac{dy}{dx}=\cos u·\left(12x^3+5\right)$

Substitute for $u$ using the same substitution that was made at the start

$\frac{dy}{dx}=\cos \left(3x^4+5x\right)·\left(12x^3+5\right)$

$f^{\text{'}}\left(x\right)=\left(12x^3+5\right)\cos \left(3x^4+5x\right)$

(b)

Substitute the inside function for u (note here that $\tan x$ is 'inside' the exponential function)

$u=\tan x$

$y=e^u$

Differentiate $u$ with respect to $x$ and differentiate $y$ with respect to $u$

$\frac{du}{dx}={\sec }^{2}x$

$\frac{dy}{du}=e^u$

Apply the chain rule which states $\frac{dy}{dx}=\frac{dy}{du}·\frac{du}{dx}$

$\frac{dy}{dx}=e^u·{\sec }^{2}x$

Substitute for $u$ using the same substitution that was made at the start

$\frac{dy}{dx}=e^{\tan x}·{\sec }^{2}x$

$g^{\text{'}}\left(x\right)=e^{\tan x}{\sec }^{2}x$

Find the derivatives of the following functions.

(a) $f\left(x\right)=\cos x·e^{4x^3+3x^2+2}$

(b) $g\left(x\right)=\sin \left({\left(3x^4+5x^2\right)}^5\right)$

Answer:

(a)

Applying the product rule

Let $u=\cos x$ and $v=e^{4x^3+3x^2+2}$

Differentiate $u$

$u^{\text{'}}=-\sin x$

To differentiate $v$, use the chain rule
It is a good idea to use a different letter for the substitution if you have already used $u$

Let $w=4x^3+3x^2+2$, so $v=e^w$

$\frac{dw}{dx}=12x^2+6x$ and $\frac{dv}{dw}=e^w$

$\begin{array}{rcl}\frac{dv}{dx}& =& \frac{dv}{dw}·\frac{dw}{dx}\\ \frac{dv}{dx}& =& e^w·\left(12x^2+6x\right)\\ v^{\text{'}}& =& \left(12x^2+6x\right)e^{4x^3+3x^2+2}\end{array}$

This might be a chain rule application that you do not need to work out formally
The exponential term is simply multiplied by the derivative of the power

Apply the product rule, $u\text{'}v+uv\text{'}$

$f^{\text{'}}\left(x\right)=-\sin x·e^{4x^3+3x^2+2}+\cos x·\left(12x^2+6x\right)e^{4x^3+3x^2+2}$

This could be factored

$f^{\text{'}}\left(x\right)=e^{4x^3+3x^2+2}\left(\left(12x^2+6x\right)\cos x-\sin x\right)$

(b)

Make a substitution for the function inside the sin function

$u={\left(3x^4+5x^2\right)}^5$ so $y=\sin u$

The function $u$ requires the chain rule to be differentiate it
This might be a chain rule application that you do not need to work out formally, but we will show the full working for this example

Make another substitution, with a different letter, for the function inside $u$

$v=3x^4+5x^2$ so $u=v^5$

Find the derivatives of the expressions which no longer require the chain rule

$\frac{dy}{du}=\cos u$

$\frac{du}{dv}=5v^4$

$\frac{dv}{dx}=12x^3+10x$

Find an expression for $\frac{dy}{dx}$ involving these three derivatives

$\frac{dy}{dx}=\frac{dy}{du}·\frac{du}{dv}·\frac{dv}{dx}$

Fill in the derivatives

$\frac{dy}{dx}=\cos u·5v^4·\left(12x^3+10x\right)$

Substitute in $u$ and $v$

$\frac{dy}{dx}=\cos \left({\left(3x^4+5x^2\right)}^5\right)·5{\left(3x^4+5x^2\right)}^4·\left(12x^3+10x\right)$

$g^{\text{'}}\left(x\right)=5{\left(3x^4+5x^2\right)}^4\left(12x^3+10x\right)\cos \left({\left(3x^4+5x^2\right)}^5\right)$