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Decomposing the function gives \(\displaystyle f(u) = 2^{u}\), \(\displaystyle u = v^{2}\), and \(\displaystyle v = \sin{\left(x \right)}.\) Using the chain rule \(\displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}\). \(\displaystyle f'(x) = (2^{u} \ln{\left(2 \right)})(2 v)(\cos{\left(x \right)}) = 2 \cdot 2^{u} v \ln{\left(2 \right)} \cos{\left(x \right)}\). Substituting back in \(\displaystyle u\) and \(\displaystyle v\) gives \(\displaystyle f'(x) = 2 \cdot 2^{v^{2}} v \ln{\left(2 \right)} \cos{\left(x \right)} = 2 \cdot 2^{\sin^{2}{\left(x \right)}} \ln{\left(2 \right)} \sin{\left(x \right)} \cos{\left(x \right)}\).

Download \(\LaTeX\)

\begin{question}Find the derivative of  $f(x) = 2^{\sin^{2}{\left(x \right)}}$. 
    \soln{9cm}{Decomposing the function gives $f(u) = 2^{u}$, $u = v^{2}$, and $ v = \sin{\left(x \right)}.$ Using the chain rule $f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}$. $f'(x) = (2^{u} \ln{\left(2 \right)})(2 v)(\cos{\left(x \right)}) = 2 \cdot 2^{u} v \ln{\left(2 \right)} \cos{\left(x \right)}$. Substituting back in $u$ and $v$ gives $f'(x) = 2 \cdot 2^{v^{2}} v \ln{\left(2 \right)} \cos{\left(x \right)} = 2 \cdot 2^{\sin^{2}{\left(x \right)}} \ln{\left(2 \right)} \sin{\left(x \right)} \cos{\left(x \right)}$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the derivative of   <img class="equation_image" title=" \displaystyle f(x) = 2^{\sin^{2}{\left(x \right)}} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%202%5E%7B%5Csin%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%20" alt="LaTeX:  \displaystyle f(x) = 2^{\sin^{2}{\left(x \right)}} " data-equation-content=" \displaystyle f(x) = 2^{\sin^{2}{\left(x \right)}} " /> . </p> </p>

HTML for Canvas

<p> <p>Decomposing the function gives  <img class="equation_image" title=" \displaystyle f(u) = 2^{u} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%202%5E%7Bu%7D%20" alt="LaTeX:  \displaystyle f(u) = 2^{u} " data-equation-content=" \displaystyle f(u) = 2^{u} " /> ,  <img class="equation_image" title=" \displaystyle u = v^{2} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20v%5E%7B2%7D%20" alt="LaTeX:  \displaystyle u = v^{2} " data-equation-content=" \displaystyle u = v^{2} " /> , and  <img class="equation_image" title=" \displaystyle  v = \sin{\left(x \right)}. " src="/equation_images/%20%5Cdisplaystyle%20%20v%20%3D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D.%20" alt="LaTeX:  \displaystyle  v = \sin{\left(x \right)}. " data-equation-content=" \displaystyle  v = \sin{\left(x \right)}. " />  Using the chain rule  <img class="equation_image" title=" \displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bdf%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdv%7D%5Cfrac%7Bdv%7D%7Bdx%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx} " data-equation-content=" \displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx} " /> .  <img class="equation_image" title=" \displaystyle f'(x) = (2^{u} \ln{\left(2 \right)})(2 v)(\cos{\left(x \right)}) = 2 \cdot 2^{u} v \ln{\left(2 \right)} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%282%5E%7Bu%7D%20%5Cln%7B%5Cleft%282%20%5Cright%29%7D%29%282%20v%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%20%3D%202%20%5Ccdot%202%5E%7Bu%7D%20v%20%5Cln%7B%5Cleft%282%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'(x) = (2^{u} \ln{\left(2 \right)})(2 v)(\cos{\left(x \right)}) = 2 \cdot 2^{u} v \ln{\left(2 \right)} \cos{\left(x \right)} " data-equation-content=" \displaystyle f'(x) = (2^{u} \ln{\left(2 \right)})(2 v)(\cos{\left(x \right)}) = 2 \cdot 2^{u} v \ln{\left(2 \right)} \cos{\left(x \right)} " /> . Substituting back in  <img class="equation_image" title=" \displaystyle u " src="/equation_images/%20%5Cdisplaystyle%20u%20" alt="LaTeX:  \displaystyle u " data-equation-content=" \displaystyle u " />  and  <img class="equation_image" title=" \displaystyle v " src="/equation_images/%20%5Cdisplaystyle%20v%20" alt="LaTeX:  \displaystyle v " data-equation-content=" \displaystyle v " />  gives  <img class="equation_image" title=" \displaystyle f'(x) = 2 \cdot 2^{v^{2}} v \ln{\left(2 \right)} \cos{\left(x \right)} = 2 \cdot 2^{\sin^{2}{\left(x \right)}} \ln{\left(2 \right)} \sin{\left(x \right)} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%202%20%5Ccdot%202%5E%7Bv%5E%7B2%7D%7D%20v%20%5Cln%7B%5Cleft%282%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%3D%202%20%5Ccdot%202%5E%7B%5Csin%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cln%7B%5Cleft%282%20%5Cright%29%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'(x) = 2 \cdot 2^{v^{2}} v \ln{\left(2 \right)} \cos{\left(x \right)} = 2 \cdot 2^{\sin^{2}{\left(x \right)}} \ln{\left(2 \right)} \sin{\left(x \right)} \cos{\left(x \right)} " data-equation-content=" \displaystyle f'(x) = 2 \cdot 2^{v^{2}} v \ln{\left(2 \right)} \cos{\left(x \right)} = 2 \cdot 2^{\sin^{2}{\left(x \right)}} \ln{\left(2 \right)} \sin{\left(x \right)} \cos{\left(x \right)} " /> . </p> </p>