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Calculus
Derivatives
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Find the derivative of \(\displaystyle f(x) = \cos{\left(e^{x^{\frac{5}{2}}} \right)}\).

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

Download \(\LaTeX\) 

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

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the derivative of   <img class="equation_image" title=" \displaystyle f(x) = \cos{\left(e^{x^{\frac{5}{2}}} \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Ccos%7B%5Cleft%28e%5E%7Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7D%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f(x) = \cos{\left(e^{x^{\frac{5}{2}}} \right)} " data-equation-content=" \displaystyle f(x) = \cos{\left(e^{x^{\frac{5}{2}}} \right)} " /> . </p> </p>
HTML for Canvas
<p> <p>Decomposing the function gives  <img class="equation_image" title=" \displaystyle f(u) = \cos{\left(u \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%20%5Ccos%7B%5Cleft%28u%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f(u) = \cos{\left(u \right)} " data-equation-content=" \displaystyle f(u) = \cos{\left(u \right)} " /> ,  <img class="equation_image" title=" \displaystyle u = e^{v} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20e%5E%7Bv%7D%20" alt="LaTeX:  \displaystyle u = e^{v} " data-equation-content=" \displaystyle u = e^{v} " /> , and  <img class="equation_image" title=" \displaystyle  v = x^{\frac{5}{2}}. " src="/equation_images/%20%5Cdisplaystyle%20%20v%20%3D%20x%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D.%20" alt="LaTeX:  \displaystyle  v = x^{\frac{5}{2}}. " data-equation-content=" \displaystyle  v = x^{\frac{5}{2}}. " />  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) = (- \sin{\left(u \right)})(e^{v})(\frac{5 x^{\frac{3}{2}}}{2}) = - \frac{5 x^{\frac{3}{2}} e^{v} \sin{\left(u \right)}}{2} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%28-%20%5Csin%7B%5Cleft%28u%20%5Cright%29%7D%29%28e%5E%7Bv%7D%29%28%5Cfrac%7B5%20x%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B2%7D%29%20%3D%20-%20%5Cfrac%7B5%20x%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20e%5E%7Bv%7D%20%5Csin%7B%5Cleft%28u%20%5Cright%29%7D%7D%7B2%7D%20" alt="LaTeX:  \displaystyle f'(x) = (- \sin{\left(u \right)})(e^{v})(\frac{5 x^{\frac{3}{2}}}{2}) = - \frac{5 x^{\frac{3}{2}} e^{v} \sin{\left(u \right)}}{2} " data-equation-content=" \displaystyle f'(x) = (- \sin{\left(u \right)})(e^{v})(\frac{5 x^{\frac{3}{2}}}{2}) = - \frac{5 x^{\frac{3}{2}} e^{v} \sin{\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{5 x^{\frac{3}{2}} e^{v} \sin{\left(e^{v} \right)}}{2} = - \frac{5 x^{\frac{3}{2}} e^{x^{\frac{5}{2}}} \sin{\left(e^{x^{\frac{5}{2}}} \right)}}{2} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%20%5Cfrac%7B5%20x%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20e%5E%7Bv%7D%20%5Csin%7B%5Cleft%28e%5E%7Bv%7D%20%5Cright%29%7D%7D%7B2%7D%20%3D%20-%20%5Cfrac%7B5%20x%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20e%5E%7Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7D%20%5Csin%7B%5Cleft%28e%5E%7Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7D%20%5Cright%29%7D%7D%7B2%7D%20" alt="LaTeX:  \displaystyle f'(x) = - \frac{5 x^{\frac{3}{2}} e^{v} \sin{\left(e^{v} \right)}}{2} = - \frac{5 x^{\frac{3}{2}} e^{x^{\frac{5}{2}}} \sin{\left(e^{x^{\frac{5}{2}}} \right)}}{2} " data-equation-content=" \displaystyle f'(x) = - \frac{5 x^{\frac{3}{2}} e^{v} \sin{\left(e^{v} \right)}}{2} = - \frac{5 x^{\frac{3}{2}} e^{x^{\frac{5}{2}}} \sin{\left(e^{x^{\frac{5}{2}}} \right)}}{2} " /> . </p> </p>