# The derivative as a function.

• 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 $$x=a$$.
• 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

# $$f(x)=|x|$$.

• 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 $\lim_{h\to0}\frac{|a+h|-|a|}{h}=\lim_{h\to0}\frac{-(a+h)+a}{h}=-1.$

# $$f(x)=|x|$$ continued.

• 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.

# Differentiability implies continuity.

• 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 $$x=a$$.