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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 \(\displaystyle 9 \sqrt{2} \sqrt{x} \cos{\left(y^{3} \right)} - 3 x^{2} e^{y^{3}}=-27\)
Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} - 27 \sqrt{2} \sqrt{x} y^{2} y' \sin{\left(y^{3} \right)} - 9 x^{2} y^{2} y' e^{y^{3}} - 6 x e^{y^{3}} + \frac{9 \sqrt{2} \cos{\left(y^{3} \right)}}{2 \sqrt{x}} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = \frac{- 4 x^{\frac{3}{2}} e^{y^{3}} + 3 \sqrt{2} \cos{\left(y^{3} \right)}}{6 y^{2} \left(x^{\frac{5}{2}} e^{y^{3}} + 3 \sqrt{2} x \sin{\left(y^{3} \right)}\right)}\)
\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $9 \sqrt{2} \sqrt{x} \cos{\left(y^{3} \right)} - 3 x^{2} e^{y^{3}}=-27$ \soln{9cm}{Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} - 27 \sqrt{2} \sqrt{x} y^{2} y' \sin{\left(y^{3} \right)} - 9 x^{2} y^{2} y' e^{y^{3}} - 6 x e^{y^{3}} + \frac{9 \sqrt{2} \cos{\left(y^{3} \right)}}{2 \sqrt{x}} = 0 \end{equation*} Solving for $y'$ gives $y' = \frac{- 4 x^{\frac{3}{2}} e^{y^{3}} + 3 \sqrt{2} \cos{\left(y^{3} \right)}}{6 y^{2} \left(x^{\frac{5}{2}} e^{y^{3}} + 3 \sqrt{2} x \sin{\left(y^{3} \right)}\right)}$} \end{question}
\documentclass{article} \usepackage{tikz} \usepackage{amsmath} \usepackage[margin=2cm]{geometry} \usepackage{tcolorbox} \newcounter{ExamNumber} \newcounter{questioncount} \stepcounter{questioncount} \newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}} \renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}} \newif\ifShowSolution \newcommand{\soln}[2]{% \ifShowSolution% \noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else% \vspace{#1}% \fi% }% \newcommand{\hideifShowSolution}[1]{% \ifShowSolution% % \else% #1% \fi% }% \everymath{\displaystyle} \ShowSolutiontrue \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 9 \sqrt{2} \sqrt{x} \cos{\left(y^{3} \right)} - 3 x^{2} e^{y^{3}}=-27 " src="/equation_images/%20%5Cdisplaystyle%209%20%5Csqrt%7B2%7D%20%5Csqrt%7Bx%7D%20%5Ccos%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%20-%203%20x%5E%7B2%7D%20e%5E%7By%5E%7B3%7D%7D%3D-27%20" alt="LaTeX: \displaystyle 9 \sqrt{2} \sqrt{x} \cos{\left(y^{3} \right)} - 3 x^{2} e^{y^{3}}=-27 " data-equation-content=" \displaystyle 9 \sqrt{2} \sqrt{x} \cos{\left(y^{3} \right)} - 3 x^{2} e^{y^{3}}=-27 " /> </p> </p>
<p> <p>Taking the derivative of both sides using implicit differentiation gives:
<img class="equation_image" title=" - 27 \sqrt{2} \sqrt{x} y^{2} y' \sin{\left(y^{3} \right)} - 9 x^{2} y^{2} y' e^{y^{3}} - 6 x e^{y^{3}} + \frac{9 \sqrt{2} \cos{\left(y^{3} \right)}}{2 \sqrt{x}} = 0 " src="/equation_images/%20%20-%2027%20%5Csqrt%7B2%7D%20%5Csqrt%7Bx%7D%20y%5E%7B2%7D%20y%27%20%5Csin%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%20-%209%20x%5E%7B2%7D%20y%5E%7B2%7D%20y%27%20e%5E%7By%5E%7B3%7D%7D%20-%206%20x%20e%5E%7By%5E%7B3%7D%7D%20%2B%20%5Cfrac%7B9%20%5Csqrt%7B2%7D%20%5Ccos%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%7D%7B2%20%5Csqrt%7Bx%7D%7D%20%3D%200%20%20" alt="LaTeX: - 27 \sqrt{2} \sqrt{x} y^{2} y' \sin{\left(y^{3} \right)} - 9 x^{2} y^{2} y' e^{y^{3}} - 6 x e^{y^{3}} + \frac{9 \sqrt{2} \cos{\left(y^{3} \right)}}{2 \sqrt{x}} = 0 " data-equation-content=" - 27 \sqrt{2} \sqrt{x} y^{2} y' \sin{\left(y^{3} \right)} - 9 x^{2} y^{2} y' e^{y^{3}} - 6 x e^{y^{3}} + \frac{9 \sqrt{2} \cos{\left(y^{3} \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{- 4 x^{\frac{3}{2}} e^{y^{3}} + 3 \sqrt{2} \cos{\left(y^{3} \right)}}{6 y^{2} \left(x^{\frac{5}{2}} e^{y^{3}} + 3 \sqrt{2} x \sin{\left(y^{3} \right)}\right)} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B-%204%20x%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20e%5E%7By%5E%7B3%7D%7D%20%2B%203%20%5Csqrt%7B2%7D%20%5Ccos%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%7D%7B6%20y%5E%7B2%7D%20%5Cleft%28x%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20e%5E%7By%5E%7B3%7D%7D%20%2B%203%20%5Csqrt%7B2%7D%20x%20%5Csin%7B%5Cleft%28y%5E%7B3%7D%20%5Cright%29%7D%5Cright%29%7D%20" alt="LaTeX: \displaystyle y' = \frac{- 4 x^{\frac{3}{2}} e^{y^{3}} + 3 \sqrt{2} \cos{\left(y^{3} \right)}}{6 y^{2} \left(x^{\frac{5}{2}} e^{y^{3}} + 3 \sqrt{2} x \sin{\left(y^{3} \right)}\right)} " data-equation-content=" \displaystyle y' = \frac{- 4 x^{\frac{3}{2}} e^{y^{3}} + 3 \sqrt{2} \cos{\left(y^{3} \right)}}{6 y^{2} \left(x^{\frac{5}{2}} e^{y^{3}} + 3 \sqrt{2} x \sin{\left(y^{3} \right)}\right)} " /> </p> </p>