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Calculus
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Use implicit differentiation where \(\displaystyle y\) is a function of \(\displaystyle x\) to find the derivative of \(\displaystyle 9 \sqrt{7} \sqrt{y} \sin{\left(x^{3} \right)} - \cos{\left(x^{3} \right)} \cos{\left(y^{3} \right)}=5\)

Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} 27 \sqrt{7} x^{2} \sqrt{y} \cos{\left(x^{3} \right)} + 3 x^{2} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 3 y^{2} y' \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + \frac{9 \sqrt{7} y' \sin{\left(x^{3} \right)}}{2 \sqrt{y}} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = - \frac{2 x^{2} \left(\sqrt{y} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 9 \sqrt{7} y \cos{\left(x^{3} \right)}\right)}{2 y^{\frac{5}{2}} \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + 3 \sqrt{7} \sin{\left(x^{3} \right)}}\)

Download \(\LaTeX\) 

\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $9 \sqrt{7} \sqrt{y} \sin{\left(x^{3} \right)} - \cos{\left(x^{3} \right)} \cos{\left(y^{3} \right)}=5$
    \soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} 27 \sqrt{7} x^{2} \sqrt{y} \cos{\left(x^{3} \right)} + 3 x^{2} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 3 y^{2} y' \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + \frac{9 \sqrt{7} y' \sin{\left(x^{3} \right)}}{2 \sqrt{y}} = 0 \end{equation*}
Solving for $y'$ gives $y' = - \frac{2 x^{2} \left(\sqrt{y} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 9 \sqrt{7} y \cos{\left(x^{3} \right)}\right)}{2 y^{\frac{5}{2}} \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + 3 \sqrt{7} \sin{\left(x^{3} \right)}}$}

\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.}

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HTML for Canvas 
<p> <p>Use implicit differentiation where  <img class="equation_image" title=" \displaystyle y " src="/equation_images/%20%5Cdisplaystyle%20y%20" alt="LaTeX:  \displaystyle y " data-equation-content=" \displaystyle y " />  is a function of  <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX:  \displaystyle x " data-equation-content=" \displaystyle x " />  to find the derivative of  <img class="equation_image" title=" \displaystyle 9 \sqrt{7} \sqrt{y} \sin{\left(x^{3} \right)} - \cos{\left(x^{3} \right)} \cos{\left(y^{3} \right)}=5 " src="/equation_images/%20%5Cdisplaystyle%209%20%5Csqrt%7B7%7D%20%5Csqrt%7By%7D%20%5Csin%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20-%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%3D5%20" alt="LaTeX:  \displaystyle 9 \sqrt{7} \sqrt{y} \sin{\left(x^{3} \right)} - \cos{\left(x^{3} \right)} \cos{\left(y^{3} \right)}=5 " data-equation-content=" \displaystyle 9 \sqrt{7} \sqrt{y} \sin{\left(x^{3} \right)} - \cos{\left(x^{3} \right)} \cos{\left(y^{3} \right)}=5 " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative of both sides using implicit differentiation gives:
 <img class="equation_image" title="  27 \sqrt{7} x^{2} \sqrt{y} \cos{\left(x^{3} \right)} + 3 x^{2} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 3 y^{2} y' \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + \frac{9 \sqrt{7} y' \sin{\left(x^{3} \right)}}{2 \sqrt{y}} = 0  " src="/equation_images/%20%2027%20%5Csqrt%7B7%7D%20x%5E%7B2%7D%20%5Csqrt%7By%7D%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20%2B%203%20x%5E%7B2%7D%20%5Csin%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%20%2B%203%20y%5E%7B2%7D%20y%27%20%5Csin%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20%2B%20%5Cfrac%7B9%20%5Csqrt%7B7%7D%20y%27%20%5Csin%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%7D%7B2%20%5Csqrt%7By%7D%7D%20%3D%200%20%20" alt="LaTeX:   27 \sqrt{7} x^{2} \sqrt{y} \cos{\left(x^{3} \right)} + 3 x^{2} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 3 y^{2} y' \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + \frac{9 \sqrt{7} y' \sin{\left(x^{3} \right)}}{2 \sqrt{y}} = 0  " data-equation-content="  27 \sqrt{7} x^{2} \sqrt{y} \cos{\left(x^{3} \right)} + 3 x^{2} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 3 y^{2} y' \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + \frac{9 \sqrt{7} y' \sin{\left(x^{3} \right)}}{2 \sqrt{y}} = 0  " /> 
Solving for  <img class="equation_image" title=" \displaystyle y' " src="/equation_images/%20%5Cdisplaystyle%20y%27%20" alt="LaTeX:  \displaystyle y' " data-equation-content=" \displaystyle y' " />  gives  <img class="equation_image" title=" \displaystyle y' = - \frac{2 x^{2} \left(\sqrt{y} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 9 \sqrt{7} y \cos{\left(x^{3} \right)}\right)}{2 y^{\frac{5}{2}} \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + 3 \sqrt{7} \sin{\left(x^{3} \right)}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20-%20%5Cfrac%7B2%20x%5E%7B2%7D%20%5Cleft%28%5Csqrt%7By%7D%20%5Csin%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%20%2B%209%20%5Csqrt%7B7%7D%20y%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%5Cright%29%7D%7B2%20y%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Csin%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20%2B%203%20%5Csqrt%7B7%7D%20%5Csin%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%7D%20" alt="LaTeX:  \displaystyle y' = - \frac{2 x^{2} \left(\sqrt{y} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 9 \sqrt{7} y \cos{\left(x^{3} \right)}\right)}{2 y^{\frac{5}{2}} \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + 3 \sqrt{7} \sin{\left(x^{3} \right)}} " data-equation-content=" \displaystyle y' = - \frac{2 x^{2} \left(\sqrt{y} \sin{\left(x^{3} \right)} \cos{\left(y^{3} \right)} + 9 \sqrt{7} y \cos{\left(x^{3} \right)}\right)}{2 y^{\frac{5}{2}} \sin{\left(y^{3} \right)} \cos{\left(x^{3} \right)} + 3 \sqrt{7} \sin{\left(x^{3} \right)}} " /> </p> </p>