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

[latex]h^{\prime}(x)=4(2x^3+2x-1)^3(6x^{2}+2)=8(3x^{2}+1)(2x^3+2x-1)^3[/latex]

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

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

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

[latex]y=-48x-88[/latex]

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

[latex]h^{\prime}(x)=7 \cos (7x+2)[/latex]

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

[latex]h^{\prime}(x)=\frac{3-4x}{(2x+3)^4}[/latex]

Rewrite [latex]h(x)=\sin^6 (x^3)=(\sin(x^3))^6[/latex].

[latex]h^{\prime}(x)=18x^2 \sin^5 (x^3) \cos (x^3)[/latex]

Acceleration is the second derivative of position.

[latex]a(t)=-16 \sin (4t)[/latex]

[latex]28[/latex]

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

Let [latex]u=x^3[/latex].

[latex]\frac{dy}{dx}=-3x^2 \sin (x^3)[/latex]

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

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

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

Define intermediate variables:

Let [latex]u = \sin(2x^3)[/latex] and [latex]v = 2x^3[/latex]

Express [latex]y[/latex] in terms of [latex]u[/latex]:

[latex]y = \sqrt{u} = u^{\frac{1}{2}}[/latex]

Find [latex]\frac{dy}{du}[/latex]:

[latex]\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}}[/latex]

Find [latex]\frac{du}{dv}[/latex]:

[latex]\frac{du}{dv} = \cos(v)[/latex]

Find [latex]\frac{dv}{dx}[/latex]:

[latex]\frac{dv}{dx} = 6x^2[/latex]

Apply the chain rule:

[latex]\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}[/latex]

Substitute and simplify:

[latex]\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}[/latex]