The Chain Rule

With the Chain Rule, the derivatives of all sorts of strange and wonderful functions are available. If we know $\mathbf{f'}$ and $g^{\prime}$, then we also know the derivatives of their composition: $( f( g(x) ) ' = f '( g(x) ) g '(x)$.

Example 9: $\quad$ Given $\mathbf{D}(\arcsin (x))=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$, find $\mathbf{D}(\arcsin (5 \mathrm{x}))$ and $\frac{\mathbf{d}\left(\arcsin \left(\mathrm{e}^{\mathrm{x}}\right)\right)}{\mathrm{dx}}$.

Solution: (a) $\arcsin (5 x)$ is the composition of $f(x)=\arcsin (x)$ with $g(x)=5 x$. We know $g^{\prime}(x)=5$, and $\mathbf{f}^{\prime}(x)=\frac{1}{\sqrt{1-x^{2}}}$ so $\mathbf{f}^{\prime}(g(x))=\frac{1}{\sqrt{1-(g(x))^{2}}}=\frac{1}{\sqrt{1-25 x^{2}}}$.

Then $\mathbf{D}(\arcsin (5 x))=f^{\prime}(g(x)) \cdot g^{\prime}(x)=\frac{1}{\sqrt{1-25 x^{2}}} \cdot(5)=\frac{5}{\sqrt{1-25 x^{2}}}$

(b) $\mathrm{y}=\arcsin \left(\mathrm{e}^{\mathrm{x}}\right)$ is $\mathrm{y}=\arcsin (\mathrm{u})$ with $\mathrm{u}=\mathrm{e}^{\mathrm{x}}$. We know $\mathrm{dy} / \mathrm{du}=\frac{1}{\sqrt{1-\mathrm{u}^{2}}}$ and $\mathrm{du} / \mathrm{dx}=\mathrm{e}^{\mathrm{x}}$ so $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{u}^{2}}} \cdot \mathrm{e}^{\mathrm{x}}=\frac{1}{\sqrt{1-\left(\mathrm{e}^{\mathrm{X}}\right)^{2}}} \cdot \mathrm{e}^{\mathrm{x}}=\frac{\mathrm{e}^{\mathrm{x}}}{\sqrt{1-\mathrm{e}^{2 \mathrm{x}}}}$

In general, $\mathbf{D}(\arcsin (\mathrm{f}(\mathrm{x})))=\frac{\mathbf{f}^{\prime}(\mathbf{x})}{\sqrt{1-(\mathrm{f}(\mathrm{x}))^{2}}}$ and $\frac{\mathbf{d}(\arcsin (\mathrm{u}))}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{u}^{2}}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Practice 7: $\mathbf{D}(\arctan (x))=\frac{1}{1+x^{2}} .$ Find $\mathbf{D}\left(\arctan \left(x^{3}\right)\right)$ and $\frac{\mathbf{d}\left(\arctan \left(e^{x}\right)\right)}{\mathbf{d x}}$

Appendix B in the back of this book shows the derivative patterns for a variety of functions. You may not know much about some of the functions, but with the differentiation patterns given and the Chain Rule you should be able to calculate derivatives of compositions. It is just a matter of following the pattern.

Practice 8: Use the patterns $\mathbf{D}(\sinh (x))=\cosh (x)$ and $D(\ln (x))=1 / x$ to determine

(a) $\mathrm{D}(\sinh (5 x-7))$

(b) $\frac{\mathbf{d}}{\mathbf{d x}}\left(\ln \left(3+\mathrm{e}^{2 \mathrm{x}}\right)\right)$

(c) $\mathrm{D}(\arcsin (1+3 \mathrm{x}))$.

Example 10: If $D(F(x))=e^{x} \cdot \sin (x)$, find $D(F(5 x))$ and $\frac{\mathbf{d}\left(F\left(t^{3}\right)\right)}{d t}$

Solution: (a) $D(F(5 x))=D\left(F(g(x))\right.$ with $g(x)=5 x \cdot F^{\prime}(x)=e^{x} \cdot \sin (x)$ so

$\begin{align*}D(F(5 x))=F^{\prime}(g(x)) \cdot g^{\prime}(x)=e^{g(x)} \cdot \sin (g(x)) \cdot 5=e^{5 x} \cdot \sin (5 x) \cdot 5\end{align*}$

(b) $\mathrm{y}=\mathrm{F}(\mathrm{u})$ with $\mathrm{u}=\mathrm{t}^{3} \cdot \frac{\mathrm{dy}}{\mathrm{du}}=\mathrm{e}^{\mathrm{u}} \cdot \sin (\mathrm{u})$ and $\frac{\mathrm{du}}{\mathrm{dt}}=3 \mathrm{t}^{2}$ so

$\begin{align*}\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{e}^{\mathrm{u}} \cdot \sin (\mathrm{u}) \cdot 3 \mathrm{t}^{2}=\mathrm{e}^{\left(\mathrm{t}^{3}\right)} \cdot \sin \left(\mathrm{t}^{3}\right) \cdot 3 \mathrm{t}^{2}\end{align*}$