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Decomposing the function gives \(\displaystyle f(u) = \sin{\left(u \right)}\), \(\displaystyle u = v^{\frac{7}{2}}\), and \(\displaystyle v = x^{3}.\) Using the chain rule \(\displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}\). \(\displaystyle f'(x) = (\cos{\left(u \right)})(\frac{7 v^{\frac{5}{2}}}{2})(3 x^{2}) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(u \right)}}{2}\). Substituting back in \(\displaystyle u\) and \(\displaystyle v\) gives \(\displaystyle f'(x) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(v^{\frac{7}{2}} \right)}}{2} = \frac{21 x^{2} \left(x^{3}\right)^{\frac{5}{2}} \cos{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)}}{2}\).

Download \(\LaTeX\)

\begin{question}Find the derivative of  $f(x) = \sin{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)}$. 
    \soln{9cm}{Decomposing the function gives $f(u) = \sin{\left(u \right)}$, $u = v^{\frac{7}{2}}$, and $ v = x^{3}.$ Using the chain rule $f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}$. $f'(x) = (\cos{\left(u \right)})(\frac{7 v^{\frac{5}{2}}}{2})(3 x^{2}) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(u \right)}}{2}$. Substituting back in $u$ and $v$ gives $f'(x) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(v^{\frac{7}{2}} \right)}}{2} = \frac{21 x^{2} \left(x^{3}\right)^{\frac{5}{2}} \cos{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)}}{2}$. }

\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) = \sin{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csin%7B%5Cleft%28%5Cleft%28x%5E%7B3%7D%5Cright%29%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f(x) = \sin{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)} " data-equation-content=" \displaystyle f(x) = \sin{\left(\left(x^{3}\right)^{\frac{7}{2}} \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^{\frac{7}{2}} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20v%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%20" alt="LaTeX:  \displaystyle u = v^{\frac{7}{2}} " data-equation-content=" \displaystyle u = v^{\frac{7}{2}} " /> , and  <img class="equation_image" title=" \displaystyle  v = x^{3}. " src="/equation_images/%20%5Cdisplaystyle%20%20v%20%3D%20x%5E%7B3%7D.%20" alt="LaTeX:  \displaystyle  v = x^{3}. " data-equation-content=" \displaystyle  v = x^{3}. " />  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)})(\frac{7 v^{\frac{5}{2}}}{2})(3 x^{2}) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(u \right)}}{2} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%28%5Ccos%7B%5Cleft%28u%20%5Cright%29%7D%29%28%5Cfrac%7B7%20v%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7D%7B2%7D%29%283%20x%5E%7B2%7D%29%20%3D%20%5Cfrac%7B21%20v%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20x%5E%7B2%7D%20%5Ccos%7B%5Cleft%28u%20%5Cright%29%7D%7D%7B2%7D%20" alt="LaTeX:  \displaystyle f'(x) = (\cos{\left(u \right)})(\frac{7 v^{\frac{5}{2}}}{2})(3 x^{2}) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(u \right)}}{2} " data-equation-content=" \displaystyle f'(x) = (\cos{\left(u \right)})(\frac{7 v^{\frac{5}{2}}}{2})(3 x^{2}) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(u \right)}}{2} " /> . 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) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(v^{\frac{7}{2}} \right)}}{2} = \frac{21 x^{2} \left(x^{3}\right)^{\frac{5}{2}} \cos{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)}}{2} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B21%20v%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20x%5E%7B2%7D%20%5Ccos%7B%5Cleft%28v%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%20%5Cright%29%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B21%20x%5E%7B2%7D%20%5Cleft%28x%5E%7B3%7D%5Cright%29%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Ccos%7B%5Cleft%28%5Cleft%28x%5E%7B3%7D%5Cright%29%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%20%5Cright%29%7D%7D%7B2%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(v^{\frac{7}{2}} \right)}}{2} = \frac{21 x^{2} \left(x^{3}\right)^{\frac{5}{2}} \cos{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)}}{2} " data-equation-content=" \displaystyle f'(x) = \frac{21 v^{\frac{5}{2}} x^{2} \cos{\left(v^{\frac{7}{2}} \right)}}{2} = \frac{21 x^{2} \left(x^{3}\right)^{\frac{5}{2}} \cos{\left(\left(x^{3}\right)^{\frac{7}{2}} \right)}}{2} " /> . </p> </p>