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Calculus
Derivatives
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Find the derivative of \(\displaystyle f(x) = \sin{\left(\cos^{4}{\left(x \right)} \right)}\).

Decomposing the function gives \(\displaystyle f(u) = \sin{\left(u \right)}\), \(\displaystyle u = v^{4}\), and \(\displaystyle v = \cos{\left(x \right)}.\) Using the chain rule \(\displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}\). \(\displaystyle f'(x) = (\cos{\left(u \right)})(4 v^{3})(- \sin{\left(x \right)}) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(u \right)}\). Substituting back in \(\displaystyle u\) and \(\displaystyle v\) gives \(\displaystyle f'(x) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(v^{4} \right)} = - 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)} \cos{\left(\cos^{4}{\left(x \right)} \right)}\).

Download \(\LaTeX\) 

\begin{question}Find the derivative of  $f(x) = \sin{\left(\cos^{4}{\left(x \right)} \right)}$. 
    \soln{9cm}{Decomposing the function gives $f(u) = \sin{\left(u \right)}$, $u = v^{4}$, and $ v = \cos{\left(x \right)}.$ Using the chain rule $f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}$. $f'(x) = (\cos{\left(u \right)})(4 v^{3})(- \sin{\left(x \right)}) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(u \right)}$. Substituting back in $u$ and $v$ gives $f'(x) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(v^{4} \right)} = - 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)} \cos{\left(\cos^{4}{\left(x \right)} \right)}$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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\begin{document}\begin{question}(10pts) The question goes here!
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HTML for Canvas 
<p> <p>Find the derivative of   <img class="equation_image" title=" \displaystyle f(x) = \sin{\left(\cos^{4}{\left(x \right)} \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csin%7B%5Cleft%28%5Ccos%5E%7B4%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f(x) = \sin{\left(\cos^{4}{\left(x \right)} \right)} " data-equation-content=" \displaystyle f(x) = \sin{\left(\cos^{4}{\left(x \right)} \right)} " /> . </p> </p>
HTML for Canvas
<p> <p>Decomposing the function gives  <img class="equation_image" title=" \displaystyle f(u) = \sin{\left(u \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%20%5Csin%7B%5Cleft%28u%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f(u) = \sin{\left(u \right)} " data-equation-content=" \displaystyle f(u) = \sin{\left(u \right)} " /> ,  <img class="equation_image" title=" \displaystyle u = v^{4} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20v%5E%7B4%7D%20" alt="LaTeX:  \displaystyle u = v^{4} " data-equation-content=" \displaystyle u = v^{4} " /> , and  <img class="equation_image" title=" \displaystyle  v = \cos{\left(x \right)}. " src="/equation_images/%20%5Cdisplaystyle%20%20v%20%3D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D.%20" alt="LaTeX:  \displaystyle  v = \cos{\left(x \right)}. " data-equation-content=" \displaystyle  v = \cos{\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) = (\cos{\left(u \right)})(4 v^{3})(- \sin{\left(x \right)}) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(u \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%28%5Ccos%7B%5Cleft%28u%20%5Cright%29%7D%29%284%20v%5E%7B3%7D%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%20%3D%20-%204%20v%5E%7B3%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28u%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'(x) = (\cos{\left(u \right)})(4 v^{3})(- \sin{\left(x \right)}) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(u \right)} " data-equation-content=" \displaystyle f'(x) = (\cos{\left(u \right)})(4 v^{3})(- \sin{\left(x \right)}) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(u \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) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(v^{4} \right)} = - 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)} \cos{\left(\cos^{4}{\left(x \right)} \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%204%20v%5E%7B3%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28v%5E%7B4%7D%20%5Cright%29%7D%20%3D%20-%204%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%5E%7B3%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28%5Ccos%5E%7B4%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'(x) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(v^{4} \right)} = - 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)} \cos{\left(\cos^{4}{\left(x \right)} \right)} " data-equation-content=" \displaystyle f'(x) = - 4 v^{3} \sin{\left(x \right)} \cos{\left(v^{4} \right)} = - 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)} \cos{\left(\cos^{4}{\left(x \right)} \right)} " /> . </p> </p>