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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 \(\displaystyle y = \cos{\left(x y \right)}\)
Taking the derivative gives \(\displaystyle y' = - \sin{\left(x y \right)}\cdot (xy'+y)\). Solving for \(\displaystyle y'\) gives \(\displaystyle y' = \left\{- \frac{y \sin{\left(x y \right)}}{x \sin{\left(x y \right)} + 1}\right\}\)
\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $y = \cos{\left(x y \right)}$
\soln{9cm}{Taking the derivative gives $y' = - \sin{\left(x y \right)}\cdot (xy'+y)$. Solving for $y'$ gives $y' = \left\{- \frac{y \sin{\left(x y \right)}}{x \sin{\left(x y \right)} + 1}\right\}$}
\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 y = \cos{\left(x y \right)} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%5Ccos%7B%5Cleft%28x%20y%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle y = \cos{\left(x y \right)} " data-equation-content=" \displaystyle y = \cos{\left(x y \right)} " /> </p> </p><p> <p>Taking the derivative gives <img class="equation_image" title=" \displaystyle y' = - \sin{\left(x y \right)}\cdot (xy'+y) " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20-%20%5Csin%7B%5Cleft%28x%20y%20%5Cright%29%7D%5Ccdot%20%28xy%27%2By%29%20" alt="LaTeX: \displaystyle y' = - \sin{\left(x y \right)}\cdot (xy'+y) " data-equation-content=" \displaystyle y' = - \sin{\left(x y \right)}\cdot (xy'+y) " /> . 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' = \left\{- \frac{y \sin{\left(x y \right)}}{x \sin{\left(x y \right)} + 1}\right\} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Cleft%5C%7B-%20%5Cfrac%7By%20%5Csin%7B%5Cleft%28x%20y%20%5Cright%29%7D%7D%7Bx%20%5Csin%7B%5Cleft%28x%20y%20%5Cright%29%7D%20%2B%201%7D%5Cright%5C%7D%20" alt="LaTeX: \displaystyle y' = \left\{- \frac{y \sin{\left(x y \right)}}{x \sin{\left(x y \right)} + 1}\right\} " data-equation-content=" \displaystyle y' = \left\{- \frac{y \sin{\left(x y \right)}}{x \sin{\left(x y \right)} + 1}\right\} " /> </p> </p>