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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 \(\displaystyle 3 x^{3} \sqrt{y} - 7 e^{y} \log{\left(x \right)}=37\)

Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} \frac{3 x^{3} y'}{2 \sqrt{y}} + 9 x^{2} \sqrt{y} - 7 y' e^{y} \log{\left(x \right)} - \frac{7 e^{y}}{x} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = \frac{2 \left(- 9 x^{3} y + 7 \sqrt{y} e^{y}\right)}{x \left(3 x^{3} - 14 \sqrt{y} e^{y} \log{\left(x \right)}\right)}\)

Download \(\LaTeX\) 

\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $3 x^{3} \sqrt{y} - 7 e^{y} \log{\left(x \right)}=37$
    \soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} \frac{3 x^{3} y'}{2 \sqrt{y}} + 9 x^{2} \sqrt{y} - 7 y' e^{y} \log{\left(x \right)} - \frac{7 e^{y}}{x} = 0 \end{equation*}
Solving for $y'$ gives $y' = \frac{2 \left(- 9 x^{3} y + 7 \sqrt{y} e^{y}\right)}{x \left(3 x^{3} - 14 \sqrt{y} e^{y} \log{\left(x \right)}\right)}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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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 3 x^{3} \sqrt{y} - 7 e^{y} \log{\left(x \right)}=37 " src="/equation_images/%20%5Cdisplaystyle%203%20x%5E%7B3%7D%20%5Csqrt%7By%7D%20-%207%20e%5E%7By%7D%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%3D37%20" alt="LaTeX:  \displaystyle 3 x^{3} \sqrt{y} - 7 e^{y} \log{\left(x \right)}=37 " data-equation-content=" \displaystyle 3 x^{3} \sqrt{y} - 7 e^{y} \log{\left(x \right)}=37 " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative of both sides using implicit differentiation gives:
 <img class="equation_image" title="  \frac{3 x^{3} y'}{2 \sqrt{y}} + 9 x^{2} \sqrt{y} - 7 y' e^{y} \log{\left(x \right)} - \frac{7 e^{y}}{x} = 0  " src="/equation_images/%20%20%5Cfrac%7B3%20x%5E%7B3%7D%20y%27%7D%7B2%20%5Csqrt%7By%7D%7D%20%2B%209%20x%5E%7B2%7D%20%5Csqrt%7By%7D%20-%207%20y%27%20e%5E%7By%7D%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cfrac%7B7%20e%5E%7By%7D%7D%7Bx%7D%20%3D%200%20%20" alt="LaTeX:   \frac{3 x^{3} y'}{2 \sqrt{y}} + 9 x^{2} \sqrt{y} - 7 y' e^{y} \log{\left(x \right)} - \frac{7 e^{y}}{x} = 0  " data-equation-content="  \frac{3 x^{3} y'}{2 \sqrt{y}} + 9 x^{2} \sqrt{y} - 7 y' e^{y} \log{\left(x \right)} - \frac{7 e^{y}}{x} = 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 \left(- 9 x^{3} y + 7 \sqrt{y} e^{y}\right)}{x \left(3 x^{3} - 14 \sqrt{y} e^{y} \log{\left(x \right)}\right)} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B2%20%5Cleft%28-%209%20x%5E%7B3%7D%20y%20%2B%207%20%5Csqrt%7By%7D%20e%5E%7By%7D%5Cright%29%7D%7Bx%20%5Cleft%283%20x%5E%7B3%7D%20-%2014%20%5Csqrt%7By%7D%20e%5E%7By%7D%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29%7D%20" alt="LaTeX:  \displaystyle y' = \frac{2 \left(- 9 x^{3} y + 7 \sqrt{y} e^{y}\right)}{x \left(3 x^{3} - 14 \sqrt{y} e^{y} \log{\left(x \right)}\right)} " data-equation-content=" \displaystyle y' = \frac{2 \left(- 9 x^{3} y + 7 \sqrt{y} e^{y}\right)}{x \left(3 x^{3} - 14 \sqrt{y} e^{y} \log{\left(x \right)}\right)} " /> </p> </p>