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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 \(\displaystyle 8 \sqrt{x} \sqrt{y} - 4 e^{x^{3}} e^{y^{3}}=49\)

Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} \frac{4 \sqrt{x} y'}{\sqrt{y}} - 12 x^{2} e^{x^{3}} e^{y^{3}} - 12 y^{2} y' e^{x^{3}} e^{y^{3}} + \frac{4 \sqrt{y}}{\sqrt{x}} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = \frac{- 3 x^{\frac{5}{2}} \sqrt{y} e^{x^{3} + y^{3}} + y}{3 \sqrt{x} y^{\frac{5}{2}} e^{x^{3} + y^{3}} - x}\)

Download \(\LaTeX\) 

\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $8 \sqrt{x} \sqrt{y} - 4 e^{x^{3}} e^{y^{3}}=49$
    \soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} \frac{4 \sqrt{x} y'}{\sqrt{y}} - 12 x^{2} e^{x^{3}} e^{y^{3}} - 12 y^{2} y' e^{x^{3}} e^{y^{3}} + \frac{4 \sqrt{y}}{\sqrt{x}} = 0 \end{equation*}
Solving for $y'$ gives $y' = \frac{- 3 x^{\frac{5}{2}} \sqrt{y} e^{x^{3} + y^{3}} + y}{3 \sqrt{x} y^{\frac{5}{2}} e^{x^{3} + y^{3}} - x}$}

\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 8 \sqrt{x} \sqrt{y} - 4 e^{x^{3}} e^{y^{3}}=49 " src="/equation_images/%20%5Cdisplaystyle%208%20%5Csqrt%7Bx%7D%20%5Csqrt%7By%7D%20-%204%20e%5E%7Bx%5E%7B3%7D%7D%20e%5E%7By%5E%7B3%7D%7D%3D49%20" alt="LaTeX:  \displaystyle 8 \sqrt{x} \sqrt{y} - 4 e^{x^{3}} e^{y^{3}}=49 " data-equation-content=" \displaystyle 8 \sqrt{x} \sqrt{y} - 4 e^{x^{3}} e^{y^{3}}=49 " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative of both sides using implicit differentiation gives:
 <img class="equation_image" title="  \frac{4 \sqrt{x} y'}{\sqrt{y}} - 12 x^{2} e^{x^{3}} e^{y^{3}} - 12 y^{2} y' e^{x^{3}} e^{y^{3}} + \frac{4 \sqrt{y}}{\sqrt{x}} = 0  " src="/equation_images/%20%20%5Cfrac%7B4%20%5Csqrt%7Bx%7D%20y%27%7D%7B%5Csqrt%7By%7D%7D%20-%2012%20x%5E%7B2%7D%20e%5E%7Bx%5E%7B3%7D%7D%20e%5E%7By%5E%7B3%7D%7D%20-%2012%20y%5E%7B2%7D%20y%27%20e%5E%7Bx%5E%7B3%7D%7D%20e%5E%7By%5E%7B3%7D%7D%20%2B%20%5Cfrac%7B4%20%5Csqrt%7By%7D%7D%7B%5Csqrt%7Bx%7D%7D%20%3D%200%20%20" alt="LaTeX:   \frac{4 \sqrt{x} y'}{\sqrt{y}} - 12 x^{2} e^{x^{3}} e^{y^{3}} - 12 y^{2} y' e^{x^{3}} e^{y^{3}} + \frac{4 \sqrt{y}}{\sqrt{x}} = 0  " data-equation-content="  \frac{4 \sqrt{x} y'}{\sqrt{y}} - 12 x^{2} e^{x^{3}} e^{y^{3}} - 12 y^{2} y' e^{x^{3}} e^{y^{3}} + \frac{4 \sqrt{y}}{\sqrt{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{- 3 x^{\frac{5}{2}} \sqrt{y} e^{x^{3} + y^{3}} + y}{3 \sqrt{x} y^{\frac{5}{2}} e^{x^{3} + y^{3}} - x} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B-%203%20x%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Csqrt%7By%7D%20e%5E%7Bx%5E%7B3%7D%20%2B%20y%5E%7B3%7D%7D%20%2B%20y%7D%7B3%20%5Csqrt%7Bx%7D%20y%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20e%5E%7Bx%5E%7B3%7D%20%2B%20y%5E%7B3%7D%7D%20-%20x%7D%20" alt="LaTeX:  \displaystyle y' = \frac{- 3 x^{\frac{5}{2}} \sqrt{y} e^{x^{3} + y^{3}} + y}{3 \sqrt{x} y^{\frac{5}{2}} e^{x^{3} + y^{3}} - x} " data-equation-content=" \displaystyle y' = \frac{- 3 x^{\frac{5}{2}} \sqrt{y} e^{x^{3} + y^{3}} + y}{3 \sqrt{x} y^{\frac{5}{2}} e^{x^{3} + y^{3}} - x} " /> </p> </p>