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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 \(\displaystyle - 2 x e^{y^{3}} - 14 \sqrt{y} e^{x^{3}}=-19\)

Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} - 42 x^{2} \sqrt{y} e^{x^{3}} - 6 x y^{2} y' e^{y^{3}} - 2 e^{y^{3}} - \frac{7 y' e^{x^{3}}}{\sqrt{y}} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = - \frac{42 x^{2} y e^{x^{3}} + 2 \sqrt{y} e^{y^{3}}}{6 x y^{\frac{5}{2}} e^{y^{3}} + 7 e^{x^{3}}}\)

Download \(\LaTeX\) 

\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $- 2 x e^{y^{3}} - 14 \sqrt{y} e^{x^{3}}=-19$
    \soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} - 42 x^{2} \sqrt{y} e^{x^{3}} - 6 x y^{2} y' e^{y^{3}} - 2 e^{y^{3}} - \frac{7 y' e^{x^{3}}}{\sqrt{y}} = 0 \end{equation*}
Solving for $y'$ gives $y' = - \frac{42 x^{2} y e^{x^{3}} + 2 \sqrt{y} e^{y^{3}}}{6 x y^{\frac{5}{2}} e^{y^{3}} + 7 e^{x^{3}}}$}

\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 - 2 x e^{y^{3}} - 14 \sqrt{y} e^{x^{3}}=-19 " src="/equation_images/%20%5Cdisplaystyle%20-%202%20x%20e%5E%7By%5E%7B3%7D%7D%20-%2014%20%5Csqrt%7By%7D%20e%5E%7Bx%5E%7B3%7D%7D%3D-19%20" alt="LaTeX:  \displaystyle - 2 x e^{y^{3}} - 14 \sqrt{y} e^{x^{3}}=-19 " data-equation-content=" \displaystyle - 2 x e^{y^{3}} - 14 \sqrt{y} e^{x^{3}}=-19 " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative of both sides using implicit differentiation gives:
 <img class="equation_image" title="  - 42 x^{2} \sqrt{y} e^{x^{3}} - 6 x y^{2} y' e^{y^{3}} - 2 e^{y^{3}} - \frac{7 y' e^{x^{3}}}{\sqrt{y}} = 0  " src="/equation_images/%20%20-%2042%20x%5E%7B2%7D%20%5Csqrt%7By%7D%20e%5E%7Bx%5E%7B3%7D%7D%20-%206%20x%20y%5E%7B2%7D%20y%27%20e%5E%7By%5E%7B3%7D%7D%20-%202%20e%5E%7By%5E%7B3%7D%7D%20-%20%5Cfrac%7B7%20y%27%20e%5E%7Bx%5E%7B3%7D%7D%7D%7B%5Csqrt%7By%7D%7D%20%3D%200%20%20" alt="LaTeX:   - 42 x^{2} \sqrt{y} e^{x^{3}} - 6 x y^{2} y' e^{y^{3}} - 2 e^{y^{3}} - \frac{7 y' e^{x^{3}}}{\sqrt{y}} = 0  " data-equation-content="  - 42 x^{2} \sqrt{y} e^{x^{3}} - 6 x y^{2} y' e^{y^{3}} - 2 e^{y^{3}} - \frac{7 y' e^{x^{3}}}{\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{42 x^{2} y e^{x^{3}} + 2 \sqrt{y} e^{y^{3}}}{6 x y^{\frac{5}{2}} e^{y^{3}} + 7 e^{x^{3}}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20-%20%5Cfrac%7B42%20x%5E%7B2%7D%20y%20e%5E%7Bx%5E%7B3%7D%7D%20%2B%202%20%5Csqrt%7By%7D%20e%5E%7By%5E%7B3%7D%7D%7D%7B6%20x%20y%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20e%5E%7By%5E%7B3%7D%7D%20%2B%207%20e%5E%7Bx%5E%7B3%7D%7D%7D%20" alt="LaTeX:  \displaystyle y' = - \frac{42 x^{2} y e^{x^{3}} + 2 \sqrt{y} e^{y^{3}}}{6 x y^{\frac{5}{2}} e^{y^{3}} + 7 e^{x^{3}}} " data-equation-content=" \displaystyle y' = - \frac{42 x^{2} y e^{x^{3}} + 2 \sqrt{y} e^{y^{3}}}{6 x y^{\frac{5}{2}} e^{y^{3}} + 7 e^{x^{3}}} " /> </p> </p>