1. Identify the functions:
   • Outer function: $f(u) = e^u$
   • Middle function: $g(v) = \sin(v)$
   • Inner function: $k(x) = x^2$
2. Apply the chain rule iteratively: $h'(x) = f'(g(k(x))) \cdot g'(k(x)) \cdot k'(x)$
3. Compute each derivative:
   • $f'(u) = e^u$
   • $g'(v) = \cos(v)$
   • $k'(x) = 2x$
4. Substitute back into the chain rule: $h'(x) = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x$
5. Simplify: $h'(x) = 2x e^{\sin(x^2)} \cos(x^2)$

$h^{\prime}(x)=4(2x^3+2x-1)^3(6x^{2}+2)=8(3x^{2}+1)(2x^3+2x-1)^3$

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You can view the transcript for this segmented clip of “3.6 The Chain Rule” here (opens in new window).

$y=-48x-88$

Apply the chain rule to $h(x)= \sin g(x)$ first and then use $g(x)=7x+2$.

$h^{\prime}(x)=7 \cos (7x+2)$

Start out by applying the quotient rule. Remember to use the chain rule to differentiate the denominator.

$h^{\prime}(x)=\frac{3-4x}{(2x+3)^4}$

Rewrite $h(x)=\sin^6 (x^3)=(\sin(x^3))^6$.

$h^{\prime}(x)=18x^2 \sin^5 (x^3) \cos (x^3)$

Acceleration is the second derivative of position.

$a(t)=-16 \sin (4t)$

$28$

1. Identify the layers of composition:
   • Outermost: $h(u) = \ln(u)$
   • Middle: $f(v) = \sin(v)$
   • Innermost: $g(x) = e^{2x}$
2. Apply the chain rule, starting from the outside: $y' = \frac{1}{\sin(e^{2x})} \cdot \frac{d}{dx}[\sin(e^{2x})]$
3. Apply the chain rule to $\sin(e^{2x})$: $y' = \frac{1}{\sin(e^{2x})} \cdot \cos(e^{2x}) \cdot \frac{d}{dx}[e^{2x}]$
4. Differentiate $e^{2x}$: $y' = \frac{1}{\sin(e^{2x})} \cdot \cos(e^{2x}) \cdot e^{2x} \cdot 2$
5. Simplify: $y' = \frac{2e^{2x} \cos(e^{2x})}{\sin(e^{2x})}$

Let $u=x^3$.

$\frac{dy}{dx}=-3x^2 \sin (x^3)$

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.6 The Chain Rule” here (opens in new window).

Define intermediate variables:

Let $u = \sin(2x^3)$ and $v = 2x^3$

Express $y$ in terms of $u$:

$y = \sqrt{u} = u^{\frac{1}{2}}$

Find $\frac{dy}{du}$:

$\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}$

Find $\frac{du}{dv}$:

$\frac{du}{dv} = \cos(v)$

Find $\frac{dv}{dx}$:

$\frac{dv}{dx} = 6x^2$

Apply the chain rule:

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

Substitute and simplify:

$\begin{array}{rcl} \frac{dy}{dx} &=& \dfrac{1}{2\sqrt{u}} \cdot \cos(v) \cdot 6x^2 \\ &=& \dfrac{1}{2\sqrt{\sin(2x^3)}} \cdot \cos(2x^3) \cdot 6x^2 \\ &=& \dfrac{3x^2 \cos(2x^3)}{\sqrt{\sin(2x^3)}} \end{array}$