- Exercise. Find the equation of the tangent line to the curve \(y=\sqrt{6-x}\) at \(x=a\).
*Definition.*The*derivative of the function \(f(x)\)*is the function \(f'(x)\) defined as \[ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}, \] and it's defined wherever the limit exists.- Alternative notations are the following. \[ \frac{\mathrm df}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}f(x)=Df(x)=D_xf(x). \]
- To emphasize the connection between the function and the graph \(y=f(x)\), the derivative is sometimes also denoted by \(y'\).
- Note that as \(f(x)\) has domain \((-\infty,6]\).
- Note also that the domain of \(f'(x)=-\frac12(6-x)^{-1/2}\) is \((-\infty,6)\).
- This reflects that the derivative \(f'(x=a)\) exists only when \(f(x)\) is defined on an open interval around \(x=a\).
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- Note that \[ \lim_{x\to6^-}f'(x)=\lim_{x\to6^-}-\frac12(6-x)^{-1/2}=-\infty. \]
- This reflects that the rate of change of \(f(x)\) gets indefinitely small as \(x\) approaches 6 from the left.
- Exercises. 2.8. 22,23,29

- Let us now consider \(f(x)=|x|\).
- To compute the derivative \[ f'(x=a)=\lim_{h\to0}\frac{|a+h|-|a|}{h}, \] we'll need to do a case-by-case scenario because of the piecewise definition: \[ |x|=\begin{cases} x, & x\ge0,\\ -x, & x\le0. \end{cases} \]
- What we can use is that since we're computing a limit at \(h\to0\), we can assume that \(h\) is as close to zero as we want.
- In case \(a>0\), we can assume \(h>-a\), so that as \(a+h>0\), we get \(|a+h|=a+h\) and \(|a|=a\), and thus \[ \lim_{h\to0}\frac{|a+h|-|a|}{h}=\lim_{h\to0}\frac{(a+h)-a}{h}=1. \]
- In case \(a<0\), we can assume \(h<-a\), so that as \(a+h<0\), we get \(|a+h|=-(a+h)\) and \(|a|=-a\), and thus \[ \lim_{h\to0}\frac{|a+h|-|a|}{h}=\lim_{h\to0}\frac{-(a+h)+a}{h}=-1. \]

- In case \(a=0\), what we can do is to calculate one-sided limits.
- If \(h\to0^+\), then in particular \(h>0\), which implies \(|h|=h\) so that \[ \lim_{h\to0^+}\frac{|h|}{h}=1. \]
- If \(h\to0^-\), then in particular \(h<0\), which implies \(|h|=-h\), so that \[ \lim_{h\to0^-}\frac{|h|}{h}=-1. \]
- Note that as the one-sided limits don't agree, the two-sided limit does not exist, therefore \(f'(x)\) is not defined at \(x=0\).
- We can give the following piecewise description. \[ f'(x)=\begin{cases} 1, & x>0,\\ -1, & x<0. \end{cases} \]
- Exercises. 2.8. 60, 62.

*Theorem.*Let \(f(x)\) be a function. Suppose that \(f(x)\) is differentiable at \(x=a\). Then \(f(x)\) is continuous at \(x=a\).*Proof.*Since \(f(x)\) is differentiable at \(x=a\), the limit \[ \lim_{x\to a}\frac{f(x)-f(a)}{x-a} \] exists.- This can be used to show the following. \[ \lim_{x\to a}(f(x)-f(a))=\lim_{x\to a}(f(x)-f(a))\frac{x-a}{x-a}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\lim_{x\to a}(x-a)=0. \]
- That in turn can be used in the following. \[ \lim_{x\to a}f(x)=\lim_{x\to a}(f(x)-f(a)+f(a))=\lim_{x\to a}(f(x)-f(a))+\lim_{x\to a}f(a)=f(a). \]
- This shows that \(f(x)\) is continuous at \(x=a\).

- Note that as we've seen in the case of \(f(x)=|x|\), it is not true that if \(f(x)\) is continuous at \(x=a\), then \(f(x)\) is differentiable at \(x=a\).