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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 \(\displaystyle 9 \sqrt{x} \cos{\left(y \right)} - 9 \sqrt{y} e^{x}=45\)

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

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\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $9 \sqrt{x} \cos{\left(y \right)} - 9 \sqrt{y} e^{x}=45$
    \soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} - 9 \sqrt{x} y' \sin{\left(y \right)} - 9 \sqrt{y} e^{x} - \frac{9 y' e^{x}}{2 \sqrt{y}} + \frac{9 \cos{\left(y \right)}}{2 \sqrt{x}} = 0 \end{equation*}
Solving for $y'$ gives $y' = \frac{- 2 \sqrt{x} y e^{x} + \sqrt{y} \cos{\left(y \right)}}{\sqrt{x} e^{x} + 2 x \sqrt{y} \sin{\left(y \right)}}$}

\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 9 \sqrt{x} \cos{\left(y \right)} - 9 \sqrt{y} e^{x}=45 " src="/equation_images/%20%5Cdisplaystyle%209%20%5Csqrt%7Bx%7D%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%20-%209%20%5Csqrt%7By%7D%20e%5E%7Bx%7D%3D45%20" alt="LaTeX:  \displaystyle 9 \sqrt{x} \cos{\left(y \right)} - 9 \sqrt{y} e^{x}=45 " data-equation-content=" \displaystyle 9 \sqrt{x} \cos{\left(y \right)} - 9 \sqrt{y} e^{x}=45 " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative of both sides using implicit differentiation gives:
 <img class="equation_image" title="  - 9 \sqrt{x} y' \sin{\left(y \right)} - 9 \sqrt{y} e^{x} - \frac{9 y' e^{x}}{2 \sqrt{y}} + \frac{9 \cos{\left(y \right)}}{2 \sqrt{x}} = 0  " src="/equation_images/%20%20-%209%20%5Csqrt%7Bx%7D%20y%27%20%5Csin%7B%5Cleft%28y%20%5Cright%29%7D%20-%209%20%5Csqrt%7By%7D%20e%5E%7Bx%7D%20-%20%5Cfrac%7B9%20y%27%20e%5E%7Bx%7D%7D%7B2%20%5Csqrt%7By%7D%7D%20%2B%20%5Cfrac%7B9%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%7D%7B2%20%5Csqrt%7Bx%7D%7D%20%3D%200%20%20" alt="LaTeX:   - 9 \sqrt{x} y' \sin{\left(y \right)} - 9 \sqrt{y} e^{x} - \frac{9 y' e^{x}}{2 \sqrt{y}} + \frac{9 \cos{\left(y \right)}}{2 \sqrt{x}} = 0  " data-equation-content="  - 9 \sqrt{x} y' \sin{\left(y \right)} - 9 \sqrt{y} e^{x} - \frac{9 y' e^{x}}{2 \sqrt{y}} + \frac{9 \cos{\left(y \right)}}{2 \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{- 2 \sqrt{x} y e^{x} + \sqrt{y} \cos{\left(y \right)}}{\sqrt{x} e^{x} + 2 x \sqrt{y} \sin{\left(y \right)}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B-%202%20%5Csqrt%7Bx%7D%20y%20e%5E%7Bx%7D%20%2B%20%5Csqrt%7By%7D%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%7D%7B%5Csqrt%7Bx%7D%20e%5E%7Bx%7D%20%2B%202%20x%20%5Csqrt%7By%7D%20%5Csin%7B%5Cleft%28y%20%5Cright%29%7D%7D%20" alt="LaTeX:  \displaystyle y' = \frac{- 2 \sqrt{x} y e^{x} + \sqrt{y} \cos{\left(y \right)}}{\sqrt{x} e^{x} + 2 x \sqrt{y} \sin{\left(y \right)}} " data-equation-content=" \displaystyle y' = \frac{- 2 \sqrt{x} y e^{x} + \sqrt{y} \cos{\left(y \right)}}{\sqrt{x} e^{x} + 2 x \sqrt{y} \sin{\left(y \right)}} " /> </p> </p>