\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus

Please login to create an exam or a quiz.

Calculus
Derivatives
New Random

Use implicit differentiation where \(\displaystyle y\) is a function of \(\displaystyle x\) to find the derivative of \(\displaystyle - 15 \sqrt{x} e^{y^{2}} - 6 e^{x^{2}} \cos{\left(y \right)}=45\)

Taking the derivative of both sides using implicit differentiation gives: \begin{equation*} - 30 \sqrt{x} y y' e^{y^{2}} - 12 x e^{x^{2}} \cos{\left(y \right)} + 6 y' e^{x^{2}} \sin{\left(y \right)} - \frac{15 e^{y^{2}}}{2 \sqrt{x}} = 0 \end{equation*} Solving for \(\displaystyle y'\) gives \(\displaystyle y' = \frac{8 x^{\frac{3}{2}} e^{x^{2}} \cos{\left(y \right)} + 5 e^{y^{2}}}{4 \left(\sqrt{x} e^{x^{2}} \sin{\left(y \right)} - 5 x y e^{y^{2}}\right)}\)

Download \(\LaTeX\) 

\begin{question}Use implicit differentiation where $y$ is a function of $x$ to find the derivative of $- 15 \sqrt{x} e^{y^{2}} - 6 e^{x^{2}} \cos{\left(y \right)}=45$
    \soln{9cm}{Taking the derivative of both sides using implicit differentiation gives:
\begin{equation*} - 30 \sqrt{x} y y' e^{y^{2}} - 12 x e^{x^{2}} \cos{\left(y \right)} + 6 y' e^{x^{2}} \sin{\left(y \right)} - \frac{15 e^{y^{2}}}{2 \sqrt{x}} = 0 \end{equation*}
Solving for $y'$ gives $y' = \frac{8 x^{\frac{3}{2}} e^{x^{2}} \cos{\left(y \right)} + 5 e^{y^{2}}}{4 \left(\sqrt{x} e^{x^{2}} \sin{\left(y \right)} - 5 x y e^{y^{2}}\right)}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
\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}
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 - 15 \sqrt{x} e^{y^{2}} - 6 e^{x^{2}} \cos{\left(y \right)}=45 " src="/equation_images/%20%5Cdisplaystyle%20-%2015%20%5Csqrt%7Bx%7D%20e%5E%7By%5E%7B2%7D%7D%20-%206%20e%5E%7Bx%5E%7B2%7D%7D%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%3D45%20" alt="LaTeX:  \displaystyle - 15 \sqrt{x} e^{y^{2}} - 6 e^{x^{2}} \cos{\left(y \right)}=45 " data-equation-content=" \displaystyle - 15 \sqrt{x} e^{y^{2}} - 6 e^{x^{2}} \cos{\left(y \right)}=45 " /> </p> </p>
HTML for Canvas
<p> <p>Taking the derivative of both sides using implicit differentiation gives:
 <img class="equation_image" title="  - 30 \sqrt{x} y y' e^{y^{2}} - 12 x e^{x^{2}} \cos{\left(y \right)} + 6 y' e^{x^{2}} \sin{\left(y \right)} - \frac{15 e^{y^{2}}}{2 \sqrt{x}} = 0  " src="/equation_images/%20%20-%2030%20%5Csqrt%7Bx%7D%20y%20y%27%20e%5E%7By%5E%7B2%7D%7D%20-%2012%20x%20e%5E%7Bx%5E%7B2%7D%7D%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%20%2B%206%20y%27%20e%5E%7Bx%5E%7B2%7D%7D%20%5Csin%7B%5Cleft%28y%20%5Cright%29%7D%20-%20%5Cfrac%7B15%20e%5E%7By%5E%7B2%7D%7D%7D%7B2%20%5Csqrt%7Bx%7D%7D%20%3D%200%20%20" alt="LaTeX:   - 30 \sqrt{x} y y' e^{y^{2}} - 12 x e^{x^{2}} \cos{\left(y \right)} + 6 y' e^{x^{2}} \sin{\left(y \right)} - \frac{15 e^{y^{2}}}{2 \sqrt{x}} = 0  " data-equation-content="  - 30 \sqrt{x} y y' e^{y^{2}} - 12 x e^{x^{2}} \cos{\left(y \right)} + 6 y' e^{x^{2}} \sin{\left(y \right)} - \frac{15 e^{y^{2}}}{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{8 x^{\frac{3}{2}} e^{x^{2}} \cos{\left(y \right)} + 5 e^{y^{2}}}{4 \left(\sqrt{x} e^{x^{2}} \sin{\left(y \right)} - 5 x y e^{y^{2}}\right)} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B8%20x%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20e%5E%7Bx%5E%7B2%7D%7D%20%5Ccos%7B%5Cleft%28y%20%5Cright%29%7D%20%2B%205%20e%5E%7By%5E%7B2%7D%7D%7D%7B4%20%5Cleft%28%5Csqrt%7Bx%7D%20e%5E%7Bx%5E%7B2%7D%7D%20%5Csin%7B%5Cleft%28y%20%5Cright%29%7D%20-%205%20x%20y%20e%5E%7By%5E%7B2%7D%7D%5Cright%29%7D%20" alt="LaTeX:  \displaystyle y' = \frac{8 x^{\frac{3}{2}} e^{x^{2}} \cos{\left(y \right)} + 5 e^{y^{2}}}{4 \left(\sqrt{x} e^{x^{2}} \sin{\left(y \right)} - 5 x y e^{y^{2}}\right)} " data-equation-content=" \displaystyle y' = \frac{8 x^{\frac{3}{2}} e^{x^{2}} \cos{\left(y \right)} + 5 e^{y^{2}}}{4 \left(\sqrt{x} e^{x^{2}} \sin{\left(y \right)} - 5 x y e^{y^{2}}\right)} " /> </p> </p>