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\title[]{MA140-Engineering Calculus}
\author[ Lecture 9 ]{\large{ Lecture 9} 
\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}
The change in distance $=f(x_{0}+h)-f(x_{0})$\\
The change in time $=(x_{0}+h)-x_{0}=h$\\
The average speed between $P$ and $Q$ is $$\frac{f(x_{0}+h)-f(x_{0})}{h}$$
\begin{center}
			\includegraphics[scale=0.25]{Figures/lec60.pdf}
		\end{center}
\end{frame}
\begin{frame}
The slope of the curve $y=f(x)$ at the point $P(x_{0},f(x_{0}))$ is the number $$\lim_{h \to 0} \frac{f(x_{0}+h)-f(x_{0})}{h}$$
We call this limit the derivative of $f$ at $x_{0}$.
\begin{definition}
The derivative of the function $f(x)$ with respect to the variable $x$ is the function $f'$ or $\frac{df}{dx}$ whose value at $x$ is $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
\end{definition}
\end{frame}
\begin{frame}
\begin{example}
Use the above definition to find the derivative of $f(x)=x^{2}$
\end{example}
We know that $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ 
Here $f(x+h)=(x+h)^{2}=x^{2}+h^{2}+2hx$ so $$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0} \frac{(x^{2}+h^{2}+2hx)-x^{2}}{h}$$
$$\hspace{-1cm} =\lim_{h \to 0} \frac{h(h+2x)}{h}=\lim_{h \to 0} (h+2x)=2x$$
\end{frame}















\end{document}