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\title[]{MA140-Engineering Calculus}
\author[ Lecture 11 ]{\large{Lecture 11} 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}
\begin{frame}
\titlepage
\end{frame}


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\begin{example}
Find $dy/dx$ for $$y=\tan ^{3}[\sin ^{2}(x^{4})]$$ 
\end{example}
$y=u^{3}$\\
$u=\tan [\sin ^{2}(x^{4})]$  $=\tan (v)$\\
$v=\sin ^{2}(x^{4})$  $=t^{2}$\\
$t=\sin (x^{4})$ $=\sin (r)$\\
$r=x^{4}$\\
By the chain rule :\\
$\displaystyle{ \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dt} \cdot \frac{dt}{dr} \cdot \frac{dr}{dx}}$\\
$\displaystyle{\frac{dy}{dx}=(3u^{2})(sec^{2}(v))(2t)(\cos r)(4x^3)}$ \\
$\displaystyle{\frac{dy}{dx}}=(3(\tan ^{2}[\sin ^{2}(x^{4})])) \cdot (\sec ^{2}(\sin ^{2}(x^{4}))) \cdot (2\sin (x^{4})) \cdot (\cos x^{4}) \cdot (4x^{3})$
\end{frame}
\begin{frame}{Differentiation of Inverse functions}

It is often useful to be able to express the derivative of an inverse function in terms of the derivatives of $f$.
\begin{definition}
If $y=f^{-1}(x)$, then $x=f(y)$ and also $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{f'(y)}$$
\end{definition}


\end{frame}
\begin{frame}
\begin{example}
Let us use the inverse rule to find $\frac{dy}{dx}$, when $y=x^{1/3}$
\end{example}
{\textcolor {red} {\textit{Note:}}} We know that the answer is $\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{\frac{-2}{3}}$, so we do not have to use the inverse rule but here we aim to use this rule to differentiate the function. \\
If $y=x^{\frac{1}{3}}$, then $y^{3}=x$, or $x=y^{3}$, so $$\frac{dx}{dy}=3y^{2}$$
By the inverse rule:$$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{3y^{2}}$$
But $y=x^{\frac{1}{3}}$ so $$\frac{dy}{dx}=\frac{1}{3(x^{\frac{1}{3}})^{2}}=\frac{1}{3}x^{-\frac{2}{3}}$$ 
\end{frame}
\begin{frame}
\begin{example}
Find $dy/dx$ for  $$y=\sin ^{-1} x$$
\end{example}
Let $y=\sin ^{-1} x$, then $  x=\sin y \enspace (\textcolor{red} {\star})$,  so $$\frac{dx}{dy}=\cos y \enspace (\textcolor{red} {\star \star})$$
As $\sin^{2} y + \cos^{2} y=1$, then $\cos y=\sqrt{1-\sin^{2} y}$, so by using $(\textcolor{red} {\star})$  $$\cos y=\sqrt{1-x^{2}}$$
Now using the inverse rule and from $(\textcolor{red} {\star \star})$, we have $$\frac{dy}{dx}=\frac{1}{\cos y} \quad or \quad \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}$$   
\end{frame}
\begin{frame}{Exercises}
\begin{itemize}
\item Show that: $$\frac{d}{dx}(\cos^{-1} x)=\frac{-1}{\sqrt{1-x^{2}}}$$
\item Using the identity $1+\tan^{2} A=\sec^{2} A$, show that: 
$$\frac{d}{dx} (\tan^{-1} x)=\frac{1}{1+x^{2}}$$
\item Find $$\frac{d}{dx}[\tan^{-1} (e^{x^{2}})]$$
\end{itemize}
\end{frame}
\begin{frame}
We had $$\frac{d}{dx} (e^{x})=e^{x}$$
We will find $\frac{dy}{dx}$ when $y=\ln(x)$\\
If $y=\ln(x)$, then $x=e^{y}$, so $\frac{dx}{dy}=e^{y}$. \\
Using the inverse rule we get $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{e^{y}}=\frac{1}{x}$$ Therefore $$\frac{d}{dx}(\ln(x))=\frac{1}{x}$$
\end{frame}
\begin{frame}
A function $f(x)$ can be written in a unique way as the sum of one even function and one odd function. The decomposition is 
\begin{center}
			\includegraphics[scale=0.2]{Figures/lec80.pdf}
		\end{center} 
 
	\begin{center}
			\includegraphics[scale=0.15]{Figures/lec81.pdf}
		\end{center} 	
\end{frame}
\begin{frame}{Hyperbolic Functions}
The hyperbolic functions are defined as follows 
\begin{itemize}
\item \textit{Hyperbolic sine of $x$} $$\sinh (x)=\frac{e^{x}-e^{-x}}{2}$$
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec83.pdf}
		\end{center} 
\end{itemize}

\end{frame}
\begin{frame}
\begin{itemize}
\item \textit{Hyperbolic cosine of $x$} $$\cosh (x)=\frac{e^{x}+e^{-x}}{2}$$
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec84.pdf}
		\end{center} 
\end{itemize}
\end{frame}
\begin{frame}
\begin{itemize}
\item \textit{Hyperbolic tangent} $$\tanh (x)=\frac{\sinh (x)}{\cosh (x)}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$
\item \textit{Hyperbolic cotangent} $$\coth (x)=\frac{\cosh (x)}{\sinh (x)}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}$$ 
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec85.pdf}
		\end{center} 
\end{itemize}
\end{frame}
\begin{frame}
\begin{example}
Find $$\frac{d \sinh (x)}{dx}$$
\end{example}
We know that $$\sinh (x)=\frac{e^{x}-e^{-x}}{2}$$  So $$\frac{d \sinh (x)}{dx}=\frac{d}{dx}(\frac{e^{x}-e^{-x}}{2})=\frac{1}{2}\frac{d(e^{x}-e^{-x})}{dx}=\frac{1}{2}[e^{x}+e^{-x}]=\cosh (x)$$
\end{frame}
\begin{frame}
\begin{itemize}
\item \textit{Hyperbolic secant} $$\sech (x)=\frac{1}{\cosh (x)}=\frac{2}{e^{x}+e^{-x}}$$
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec86.pdf}
		\end{center} 
\end{itemize}
\end{frame}
\begin{frame}
\begin{itemize}
\item \textit{Hyperbolic cosecant} $$\csch (x)=\frac{1}{\sinh (x)}=\frac{2}{e^{x}-e^{-x}}$$
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec87.pdf}
		\end{center} 
\end{itemize}
\end{frame}
\begin{frame}{Exercises}
\begin{itemize}
\item Show that $$\frac{d}{dx}(\cosh (x))=\sinh (x)$$
\item Show that $$\frac{d}{dx}(\tanh (x))=\sech ^{2}(x)$$ 
\end{itemize}
\end{frame}



\end{document}