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Questions: Algebra BusinessCalculus
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Use implicit differentiation where \(\displaystyle y\) is a function of \(\displaystyle x\) to find the derivative of \(\displaystyle - 6 \sqrt{x} \sin{\left(y \right)} + \cos{\left(x \right)}=-45\)
Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} - 6 \sqrt{x} y' \cos{\left(y \right)} - \sin{\left(x \right)} - \frac{3 \sin{\left(y \right)}}{\sqrt{x}} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = - \frac{\tan{\left(y \right)}}{2 x} - \frac{\sin{\left(x \right)}}{6 \sqrt{x} \cos{\left(y \right)}}\)
\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $- 6 \sqrt{x} \sin{\left(y \right)} + \cos{\left(x \right)}=-45$
\soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} - 6 \sqrt{x} y' \cos{\left(y \right)} - \sin{\left(x \right)} - \frac{3 \sin{\left(y \right)}}{\sqrt{x}} = 0 \end{equation*}
Solving for $y'$ gives $y' = - \frac{\tan{\left(y \right)}}{2 x} - \frac{\sin{\left(x \right)}}{6 \sqrt{x} \cos{\left(y \right)}}$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<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 - 6 \sqrt{x} \sin{\left(y \right)} + \cos{\left(x \right)}=-45 " src="/equation_images/%20%5Cdisplaystyle%20-%206%20%5Csqrt%7Bx%7D%20%5Csin%7B%5Cleft%28y%20%5Cright%29%7D%20%2B%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%3D-45%20" alt="LaTeX: \displaystyle - 6 \sqrt{x} \sin{\left(y \right)} + \cos{\left(x \right)}=-45 " data-equation-content=" \displaystyle - 6 \sqrt{x} \sin{\left(y \right)} + \cos{\left(x \right)}=-45 " /> </p> </p><p> <p>Taking the derivative of both sides using implicit differentiation gives:
<img class="equation_image" title=" - 6 \sqrt{x} y' \cos{\left(y \right)} - \sin{\left(x \right)} - \frac{3 \sin{\left(y \right)}}{\sqrt{x}} = 0 " src="/equation_images/%20%20-%206%20%5Csqrt%7Bx%7D%20y%27%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%20-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cfrac%7B3%20%5Csin%7B%5Cleft%28y%20%5Cright%29%7D%7D%7B%5Csqrt%7Bx%7D%7D%20%3D%200%20%20" alt="LaTeX: - 6 \sqrt{x} y' \cos{\left(y \right)} - \sin{\left(x \right)} - \frac{3 \sin{\left(y \right)}}{\sqrt{x}} = 0 " data-equation-content=" - 6 \sqrt{x} y' \cos{\left(y \right)} - \sin{\left(x \right)} - \frac{3 \sin{\left(y \right)}}{\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{\tan{\left(y \right)}}{2 x} - \frac{\sin{\left(x \right)}}{6 \sqrt{x} \cos{\left(y \right)}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20-%20%5Cfrac%7B%5Ctan%7B%5Cleft%28y%20%5Cright%29%7D%7D%7B2%20x%7D%20-%20%5Cfrac%7B%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B6%20%5Csqrt%7Bx%7D%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%7D%20" alt="LaTeX: \displaystyle y' = - \frac{\tan{\left(y \right)}}{2 x} - \frac{\sin{\left(x \right)}}{6 \sqrt{x} \cos{\left(y \right)}} " data-equation-content=" \displaystyle y' = - \frac{\tan{\left(y \right)}}{2 x} - \frac{\sin{\left(x \right)}}{6 \sqrt{x} \cos{\left(y \right)}} " /> </p> </p>