Activity 31.
For each of the following functions, find the function’s derivative. State the rule(s) you use, label relevant derivatives appropriately, and be sure to clearly identify your overall answer.
(a)
\(p(r) = 4\sqrt{r^6 + 2e^r}\)
Hint.
Use the constant multiple rule first, followed by the chain rule.
Answer.
\(p'(r) = \frac{4(6r^5 + 2e^r)}{2\sqrt{r^6 + 2e^r}}\text{.}\)
Solution.
By the constant multiple rule, \(p'(r) = 4\frac{d}{dr}[\sqrt{r^6 + 2e^r}]\text{.}\) Using the chain rule to complete the remaining derivative, we see that
\begin{equation*}
p'(r) = 4 \frac{1}{2\sqrt{r^6 + 2e^r}} \frac{d}{dr}[r^6 + 2e^r] = \frac{4(6r^5 + 2e^r)}{2\sqrt{r^6 + 2e^r}}\text{.}
\end{equation*}
(b)
\(m(v) = \sin(v^2) \cos(v^3)\)
Hint.
Observe that \(m\) is fundamentally a product of composite functions.
Answer.
\(m'(v) = -3v^2 \sin(v^2)\sin(v^3) + 2v \cos(v^3)\cos(v^2)\text{.}\)
Solution.
Observe that by the product rule, \(m'(v) = \sin(v^2) \frac{d}{dv}[\cos(v^3)] + \cos(v^3) \frac{d}{dv}[\sin(v^2)]\text{.}\) Applying the chain rule to differentiate \(\cos(v^3)\) and \(\sin(v^2)\text{,}\) we see that
\begin{equation*}
m'(v) = \sin(v^2) [-\sin(v^3) \cdot 3v^2] + \cos(v^3) [\cos(v^2) \cdot 2v] = -3v^2 \sin(v^2)\sin(v^3) + 2v \cos(v^3)\cos(v^2)\text{.}
\end{equation*}
(c)
\(h(y) = \frac{\cos(10y)}{e^{4y}+1}\)
Hint.
Note that \(h\) is a quotient of composite functions.
Answer.
\(h'(y) = \frac{(e^{4y}+1) [-10\sin(10y)] - \cos(10y) [4e^{4y}]}{(e^{4y}+1)^2}\text{.}\)
Solution.
By the quotient rule,
\begin{equation*}
h'(y) = \frac{(e^{4y}+1) \frac{d}{dy}[\cos(10y)] - \cos(10y) \frac{d}{dy}[e^{4y}+1]}{(e^{4y}+1)^2}\text{.}
\end{equation*}
Applying the chain rule to differentiate \(\cos(10y)\) and \(e^{4y}\text{,}\) it follows
\begin{equation*}
h'(y) = \frac{(e^{4y}+1) [-10\sin(10y)] - \cos(10y) [4e^{4y}]}{(e^{4y}+1)^2}\text{.}
\end{equation*}
(d)
\(s(z) = 2^{z^2 \sec (z)}\)
Hint.
The function \(s\) is a composite function with outer function \(2^z\text{.}\)
Answer.
\(s'(z) = 2^{z^2\sec(z)} \ln(2) [z^2 \sec(z)\tan(z) + \sec(z) \cdot 2z]\text{.}\)
Solution.
By the chain rule, \(s'(z) = 2^{z^2\sec(z)} \ln(2) \frac{d}{dz}[z^2 \sec(z)]\text{.}\) Then by the product rule, we find that
\begin{equation*}
s'(z) = 2^{z^2\sec(z)} \ln(2) [z^2 \sec(z)\tan(z) + \sec(z) \cdot 2z]\text{.}
\end{equation*}
(e)
\(c(x) = \sin(e^{x^2})\)
Hint.
It is possible for a function to be a composite function with more than two functions in the chain.
Answer.
\(c'(x) = \cos(e^{x^2}) [e^{x^2}\cdot 2x]\text{.}\)
Solution.
If we first apply the chain rule to the outer function (the sine function), note that
\begin{equation*}
c'(x) = \cos(e^{x^2}) \frac{d}{dx}[e^{x^2}]\text{.}
\end{equation*}
Next, we again apply the chain rule, but this time to \(e^{x^2}\text{,}\) and get
\begin{equation*}
c'(x) = \cos(e^{x^2}) [e^{x^2}\cdot 2x]\text{.}
\end{equation*}
