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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 \(\displaystyle - 4 x \sin{\left(y^{2} \right)} + \sqrt{6} \sqrt{y} \cos{\left(x^{3} \right)}=-6\)
Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} - 3 \sqrt{6} x^{2} \sqrt{y} \sin{\left(x^{3} \right)} - 8 x y y' \cos{\left(y^{2} \right)} - 4 \sin{\left(y^{2} \right)} + \frac{\sqrt{6} y' \cos{\left(x^{3} \right)}}{2 \sqrt{y}} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = - \frac{6 \sqrt{6} x^{2} y \sin{\left(x^{3} \right)} + 8 \sqrt{y} \sin{\left(y^{2} \right)}}{16 x y^{\frac{3}{2}} \cos{\left(y^{2} \right)} - \sqrt{6} \cos{\left(x^{3} \right)}}\)
\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $- 4 x \sin{\left(y^{2} \right)} + \sqrt{6} \sqrt{y} \cos{\left(x^{3} \right)}=-6$
\soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} - 3 \sqrt{6} x^{2} \sqrt{y} \sin{\left(x^{3} \right)} - 8 x y y' \cos{\left(y^{2} \right)} - 4 \sin{\left(y^{2} \right)} + \frac{\sqrt{6} y' \cos{\left(x^{3} \right)}}{2 \sqrt{y}} = 0 \end{equation*}
Solving for $y'$ gives $y' = - \frac{6 \sqrt{6} x^{2} y \sin{\left(x^{3} \right)} + 8 \sqrt{y} \sin{\left(y^{2} \right)}}{16 x y^{\frac{3}{2}} \cos{\left(y^{2} \right)} - \sqrt{6} \cos{\left(x^{3} \right)}}$}
\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 - 4 x \sin{\left(y^{2} \right)} + \sqrt{6} \sqrt{y} \cos{\left(x^{3} \right)}=-6 " src="/equation_images/%20%5Cdisplaystyle%20-%204%20x%20%5Csin%7B%5Cleft%28y%5E%7B2%7D%20%5Cright%29%7D%20%2B%20%5Csqrt%7B6%7D%20%5Csqrt%7By%7D%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%3D-6%20" alt="LaTeX: \displaystyle - 4 x \sin{\left(y^{2} \right)} + \sqrt{6} \sqrt{y} \cos{\left(x^{3} \right)}=-6 " data-equation-content=" \displaystyle - 4 x \sin{\left(y^{2} \right)} + \sqrt{6} \sqrt{y} \cos{\left(x^{3} \right)}=-6 " /> </p> </p><p> <p>Taking the derivative of both sides using implicit differentiation gives:
<img class="equation_image" title=" - 3 \sqrt{6} x^{2} \sqrt{y} \sin{\left(x^{3} \right)} - 8 x y y' \cos{\left(y^{2} \right)} - 4 \sin{\left(y^{2} \right)} + \frac{\sqrt{6} y' \cos{\left(x^{3} \right)}}{2 \sqrt{y}} = 0 " src="/equation_images/%20%20-%203%20%5Csqrt%7B6%7D%20x%5E%7B2%7D%20%5Csqrt%7By%7D%20%5Csin%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20-%208%20x%20y%20y%27%20%5Ccos%7B%5Cleft%28y%5E%7B2%7D%20%5Cright%29%7D%20-%204%20%5Csin%7B%5Cleft%28y%5E%7B2%7D%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Csqrt%7B6%7D%20y%27%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%7D%7B2%20%5Csqrt%7By%7D%7D%20%3D%200%20%20" alt="LaTeX: - 3 \sqrt{6} x^{2} \sqrt{y} \sin{\left(x^{3} \right)} - 8 x y y' \cos{\left(y^{2} \right)} - 4 \sin{\left(y^{2} \right)} + \frac{\sqrt{6} y' \cos{\left(x^{3} \right)}}{2 \sqrt{y}} = 0 " data-equation-content=" - 3 \sqrt{6} x^{2} \sqrt{y} \sin{\left(x^{3} \right)} - 8 x y y' \cos{\left(y^{2} \right)} - 4 \sin{\left(y^{2} \right)} + \frac{\sqrt{6} y' \cos{\left(x^{3} \right)}}{2 \sqrt{y}} = 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{6 \sqrt{6} x^{2} y \sin{\left(x^{3} \right)} + 8 \sqrt{y} \sin{\left(y^{2} \right)}}{16 x y^{\frac{3}{2}} \cos{\left(y^{2} \right)} - \sqrt{6} \cos{\left(x^{3} \right)}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20-%20%5Cfrac%7B6%20%5Csqrt%7B6%7D%20x%5E%7B2%7D%20y%20%5Csin%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%20%2B%208%20%5Csqrt%7By%7D%20%5Csin%7B%5Cleft%28y%5E%7B2%7D%20%5Cright%29%7D%7D%7B16%20x%20y%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20%5Ccos%7B%5Cleft%28y%5E%7B2%7D%20%5Cright%29%7D%20-%20%5Csqrt%7B6%7D%20%5Ccos%7B%5Cleft%28x%5E%7B3%7D%20%5Cright%29%7D%7D%20" alt="LaTeX: \displaystyle y' = - \frac{6 \sqrt{6} x^{2} y \sin{\left(x^{3} \right)} + 8 \sqrt{y} \sin{\left(y^{2} \right)}}{16 x y^{\frac{3}{2}} \cos{\left(y^{2} \right)} - \sqrt{6} \cos{\left(x^{3} \right)}} " data-equation-content=" \displaystyle y' = - \frac{6 \sqrt{6} x^{2} y \sin{\left(x^{3} \right)} + 8 \sqrt{y} \sin{\left(y^{2} \right)}}{16 x y^{\frac{3}{2}} \cos{\left(y^{2} \right)} - \sqrt{6} \cos{\left(x^{3} \right)}} " /> </p> </p>