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

Please login to create an exam or a quiz.

Calculus
Derivatives
New Random

Find the derivative of \(\displaystyle y=\tan{\left(\sin{\left(x \right)} \right)}\).

The outer function is \(\displaystyle f(u) = \tan{\left(u \right)}\) and the inner function \(\displaystyle u = \sin{\left(x \right)}\). The chain rule gives \(\displaystyle y'= \frac{df}{du}\frac{du}{dx}\). \(\displaystyle y' = \tan^{2}{\left(u \right)} + 1(\cos{\left(x \right)}) = \left(\tan^{2}{\left(\sin{\left(x \right)} \right)} + 1\right) \cos{\left(x \right)}\). Simplifying further gives \(\displaystyle y'= \frac{\cos{\left(x \right)}}{\cos^{2}{\left(\sin{\left(x \right)} \right)}}\)

Download \(\LaTeX\) 

\begin{question}Find the derivative of $y=\tan{\left(\sin{\left(x \right)} \right)}$. 
    \soln{9cm}{The outer function is $f(u) = \tan{\left(u \right)}$ and the inner function $u = \sin{\left(x \right)}$. The chain rule gives $y'= \frac{df}{du}\frac{du}{dx}$. $y' = \tan^{2}{\left(u \right)} + 1(\cos{\left(x \right)}) = \left(\tan^{2}{\left(\sin{\left(x \right)} \right)} + 1\right) \cos{\left(x \right)}$. Simplifying further gives $y'= \frac{\cos{\left(x \right)}}{\cos^{2}{\left(\sin{\left(x \right)} \right)}}$}

\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>Find the derivative of  <img class="equation_image" title=" \displaystyle y=\tan{\left(\sin{\left(x \right)} \right)} " src="/equation_images/%20%5Cdisplaystyle%20y%3D%5Ctan%7B%5Cleft%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle y=\tan{\left(\sin{\left(x \right)} \right)} " data-equation-content=" \displaystyle y=\tan{\left(\sin{\left(x \right)} \right)} " /> . </p> </p>
HTML for Canvas
<p> <p>The outer function is  <img class="equation_image" title=" \displaystyle f(u) = \tan{\left(u \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%20%5Ctan%7B%5Cleft%28u%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f(u) = \tan{\left(u \right)} " data-equation-content=" \displaystyle f(u) = \tan{\left(u \right)} " />  and the inner function  <img class="equation_image" title=" \displaystyle u = \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle u = \sin{\left(x \right)} " data-equation-content=" \displaystyle u = \sin{\left(x \right)} " /> . The chain rule gives  <img class="equation_image" title=" \displaystyle y'= \frac{df}{du}\frac{du}{dx} " src="/equation_images/%20%5Cdisplaystyle%20y%27%3D%20%5Cfrac%7Bdf%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20" alt="LaTeX:  \displaystyle y'= \frac{df}{du}\frac{du}{dx} " data-equation-content=" \displaystyle y'= \frac{df}{du}\frac{du}{dx} " /> .  <img class="equation_image" title=" \displaystyle y' = \tan^{2}{\left(u \right)} + 1(\cos{\left(x \right)}) = \left(\tan^{2}{\left(\sin{\left(x \right)} \right)} + 1\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Ctan%5E%7B2%7D%7B%5Cleft%28u%20%5Cright%29%7D%20%2B%201%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%20%3D%20%5Cleft%28%5Ctan%5E%7B2%7D%7B%5Cleft%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D%20%2B%201%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle y' = \tan^{2}{\left(u \right)} + 1(\cos{\left(x \right)}) = \left(\tan^{2}{\left(\sin{\left(x \right)} \right)} + 1\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle y' = \tan^{2}{\left(u \right)} + 1(\cos{\left(x \right)}) = \left(\tan^{2}{\left(\sin{\left(x \right)} \right)} + 1\right) \cos{\left(x \right)} " /> . Simplifying further gives  <img class="equation_image" title=" \displaystyle y'= \frac{\cos{\left(x \right)}}{\cos^{2}{\left(\sin{\left(x \right)} \right)}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%3D%20%5Cfrac%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Ccos%5E%7B2%7D%7B%5Cleft%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D%7D%20" alt="LaTeX:  \displaystyle y'= \frac{\cos{\left(x \right)}}{\cos^{2}{\left(\sin{\left(x \right)} \right)}} " data-equation-content=" \displaystyle y'= \frac{\cos{\left(x \right)}}{\cos^{2}{\left(\sin{\left(x \right)} \right)}} " /> </p> </p>