Activity 30.
For each function given below, identify an inner function \(g\) and outer function \(f\) to write the function in the form \(f(g(x))\text{.}\) Determine \(f'(x)\text{,}\) \(g'(x)\text{,}\) and \(f'(g(x))\text{,}\) and then apply the chain rule to determine the derivative of the given function.
(a)
\(h(x) = \cos(x^4)\)
Hint.
The outer function is \(f(x) = \cos(x)\text{.}\)
Answer.
\(h'(x) = -4x^3\sin(x^4)\text{.}\)
Solution.
The outer function is \(f(x) = \cos(x)\text{,}\) while the inner function is \(g(x) = x^4\text{,}\) and we know that
\begin{equation*}
f'(x) = -\sin(x),
g'(x) = 4x^3, \ \text{and} \ f'(g(x)) = -\sin(x^4)\text{.}
\end{equation*}
Hence, by the chain rule,
\begin{equation*}
h'(x) = f'(g(x))g'(x) = -4x^3\sin(x^4)\text{.}
\end{equation*}
(b)
\(p(x) = \sqrt{ \tan(x) }\)
Hint.
The outer function is \(f(x) = \sqrt{x}\text{.}\)
Answer.
\(h'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\text{.}\)
Solution.
The outer function is \(f(x) = \sqrt{x}\text{,}\) while the inner function is \(g(x) = \tan(x)\text{,}\) and we know that
\begin{equation*}
f'(x) = \frac{1}{2\sqrt{x}}, g'(x) = \sec^2(x), \ \text{and} \ f'(g(x)) = \frac{1}{2\sqrt{\tan(x)}}\text{.}
\end{equation*}
Hence, by the chain rule,
\begin{equation*}
h'(x) = f'(g(x))g'(x) = \frac{\sec^2(x)}{2\sqrt{\tan(x)}}\text{.}
\end{equation*}
(c)
\(s(x) = 2^{\sin(x)}\)
Hint.
The outer function is \(f(x) = 2^x\text{.}\)
Answer.
\(h'(x) = 2^{\sin(x)}\ln(2)\cos(x)\text{.}\)
Solution.
The outer function is \(f(x) = 2^x\text{,}\) while the inner function is \(g(x) = \sin(x)\text{,}\) and we know that
\begin{equation*}
f'(x) = 2^x \ln(2),
g'(x) = \cos(x), \ \text{and} \ f'(g(x)) = 2^{\sin(x)}\ln(2)\text{.}
\end{equation*}
Hence, by the chain rule,
\begin{equation*}
h'(x) = f'(g(x))g'(x) = 2^{\sin(x)}\ln(2)\cos(x)\text{.}
\end{equation*}
(d)
\(z(x) = \cot^5(x)\)
Hint.
The outer function is \(f(x) = x^5\text{.}\)
Answer.
\(h'(x) = -5\cot^4(x) \csc^2(x)\text{.}\)
Solution.
The outer function is \(f(x) = x^5\text{,}\) while the inner function is \(g(x) = \cot(x)\text{,}\) and we know that
\begin{equation*}
f'(x) = 5x^4, g'(x) = -\csc^2(x), \ \text{and} \ f'(g(x)) = 5\cot^4(x)\text{.}
\end{equation*}
Hence, by the chain rule,
\begin{equation*}
h'(x) = f'(g(x))g'(x) = -5\cot^4(x) \csc^2(x)\text{.}
\end{equation*}
(e)
\(m(x) = (\sec(x) + e^x)^9\)
Hint.
The outer function is \(f(x) = x^9\text{.}\)
Answer.
\(h'(x) = 9(\sec(x)+e^x)^8 (\sec(x)\tan(x) + e^x)\text{.}\)
Solution.
The outer function is \(f(x) = x^9\text{,}\) while the inner function is \(g(x) = \sec(x) + e^x\text{,}\) and we know that
\begin{equation*}
f'(x) = 9x^8, g'(x) = \sec(x)\tan(x) + e^x, \ \text{and} \ f'(g(x)) = 9(\sec(x)+e^x)^8\text{.}
\end{equation*}
Hence, by the chain rule,
\begin{equation*}
h'(x) = f'(g(x))g'(x) = 9(\sec(x)+e^x)^8 (\sec(x)\tan(x) + e^x)\text{.}
\end{equation*}
