1. Identify \(f'(x)\) and \(g'(x)\) \[f'\left(x\right)=-\sin\left(x\right)\] \[g'\left(x\right)=4x\]
2. Substitute these into the formula \[\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]= -4x\sin\left(2x^{2}+1\right)\]

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Identify the outer function and inner function, and then find the first derivative.

\[f_{inner}\left(x\right)=\cos\left(x\right)\] \[f_{outer}\left(x\right)=x^{2}\]

\[f'\left(x\right)=\frac{d}{dx}\left[\cos\left(x\right)\right]\cdot2\left(\cos x\right)\]

\[f'\left(x\right)=-2\sin\left(x\right)\cos\left(x\right)\]

Find the second derivative

\[f''\left(x\right)=-2\frac{d}{dx}\left[\sin\left(x\right)\right]\cos x-2\sin\left(x\right)\cdot\frac{d}{dx}\left[\cos\left(x\right)\right]=\]

\[-2\cos\left(x\right)\cos\left(x\right)+2\sin\left(x\right)\sin\left(x\right)=2\sin^{2}\left(x\right)-2\cos^{2}\left(x\right)\]

This example is the same as example 4 from the last topic; however, when we first evaluated this example, we used the product rule twice. Now we can use the chain rule and the product rule instead.

First, use the product rule to evaluate the derivative.

\[f'\left(x\right)=\frac{d}{dx}\left[\cos^{2}\left(x\right)\right]\sin\left(x\right)+\ \frac{d}{dx}\left[\sin\left(x\right)\right]\cos^{2}\left(x\right)=\]

We will use the chain rule when evaluating the derivative of \(\cos^{2}\left(x\right)\). Like the last example, let \(\cos\left(x\right)\) be the inner function, and let \(x^{2}\) be outer function.

\[2\left(\cos\left(x\right)\right)\cdot\left(-\sin\left(x\right)\right)\cdot\sin\left(x\right)+\cos\left(x\right)\cdot\cos^{2}\left(x\right)=\]

\[-2\cos\left(x\right)\sin^{2}\left(x\right)+\cos^{3}\left(x\right)\]

\[f'\left(x\right)=3\left[\frac{d}{dx}\left[\left(2x^{3}+2\right)^{3}\right]\cos\left(x\right)+\left(2x^{3}+2\right)^{3}\cdot\frac{d}{dx}\left[\cos\left(x\right)\right]\right]\] \[3\left[\left[3\left(\frac{d}{dx}\left[2x^{3}+2\right]\right)\left(2x^{3}+2\right)^{2}\right]\cos\left(x\right)-\left(2x^{3}+2\right)^{3}\sin\left(x\right)\right]=\] \[9\left(6x^{2}\right)\left(2x^{3}+2\right)^{2}\cos\left(x\right)-\left(2x^{3}+2\right)^{3}\sin\left(x\right)\]

1. Find \(f(1)\) \[f\left(1\right)=\sqrt[3]{1+2+5}=\sqrt[3]{8}=2\]
2. Find \(f'(1)\) \[f'\left(x\right)=\left(\frac{d}{dx}\left[x^{3}+2x+5\right]\right)\left(\frac{1}{3}\right)\left(x^{3}+2x+5\right)^{\frac{1}{3}-1}=\] \[\frac{1}{3}\left(3x^{2}+2\right)\left(x^{3}+2x+5\right)^{-\frac{2}{3}}=\frac{\frac{1}{3}\left(3x^{2}+2\right)}{\sqrt[3]{\left(x^{3}+2x+5\right)^{2}}}\] \[\therefore\] \[f'\left(1\right)=\frac{\frac{1}{3}\left(5\right)}{\sqrt[3]{8^{2}}}=\frac{5}{3}\cdot\frac{1}{4}=\frac{5}{12}\]
3. Find the slope of the normal line \[m_{normal}=-\frac{1}{m_{tangent}}\] \[m_{normal}=-\frac{1}{\frac{5}{12}}=-\frac{12}{5}\]
4. Write the equation of the normal line \[y-2=\frac{5}{12}\left(x-1\right)\] \[\mbox{ or }\] \[y=-\frac{12}{5}x+\frac{12}{5}+2=-\frac{12}{5}x+\frac{22}{5}\]