$\therefore \lim _{x \rightarrow 0} f(x)=0$

Also, $f(0)=|0|=0$

$\therefore f(0)=\lim _{x \rightarrow 0} f(x)$, hence $f(x)$ is cont. at $x=0$.

Is $f(x)$ differentiable at $x=0$ ?

Does $\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$ exist?

For h ≠ 0, ${\frac{f(0+h)-f(0)}{h}}=\frac{f(h)-f(0)}{h}=\frac{|h|-0}{h}=\frac{|h|}{h}$ $={\begin{cases}1 , if h>0 \\ -1 , if h<0\end{cases}}$

$\therefore L.H.S. \neq R.H.S.$ of $~ \frac{f(0+h)-f(0)}{h}$

Hence, $\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$ does not exist.

$\therefore f(x)$ is not differentiable at $x=0$.

Conclusion: $f(x)$ continuous at $x=a$

does not imply that $f(x)$ is differentiable at $x=a$.

Theorem: If $f(x)$ is differentiable at $x=a$ then $f(x)$ must be continuous at $x=a$.

Proof: Since $f(x)$ is differentiable at $x=a$, $f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ exists.

To check the continuity at $x=a$, we need

$f(a)= \lim _{x \rightarrow a} f(x)=\lim _{h \rightarrow 0} f(a+h)$

$(\text { Put } x=a+h, \text { ie. } h=x-a,$

$\text { As } x \rightarrow a, h \rightarrow 0)$

$\lim _{h \rightarrow 0} f(a+h)=f(a) \Leftrightarrow \lim _{h \rightarrow 0}[f(a+h)-f(a)]=0$

But, for $h \neq 0, f(a+h)-f(a)=h\left[\frac{f(a+h)-f(a)}{h}\right]$

Since $\lim _{h \rightarrow 0} h=0$ and $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=f^{\prime}(a)$

$\text{ by the product rule for limits, we have}$

$\lim _{h \rightarrow 0}[f(a+h)-f(a)]=0 \cdot f^{\prime}(a)=0 \text { }$

Hence $f(x)$ is cont. at $x=a$.

$\frac{d}{d x}[f(x) g(x)]=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$

provided $f^{\prime}(x)$ & $g^{\prime}(x)$ exists.

Pf:

Let $u(x)=f(x) g(x)$.

For $h \neq 0, \quad \frac{u(x+h)-u(x)}{h}$

$=\frac{f(x+h) g(x+h)-f(x) g(x)}{h}$

$=\frac{[f(x+h) g(x+h)-f(x) g(x+h)]+[f(x) g(x+h)-f(x) g(x)]}{h}$

$\therefore \frac{u(x+h)-u(x)}{h} ={\left[\frac{f(x+h)-f(x)}{h}\right] g(x+h)}$

$+f(x)\left[\frac{g(x+h)-g(x)}{h}\right]$

Now, $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f^{\prime}(x)$

$\lim _{h \rightarrow 0} g(x+h)=g(x)$

$[\because g$ is continuous at $x$ since it is given to be differentiable at x]

similarly, $\lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=g^{\prime}(x)$

$\therefore u^{\prime}(x)=\lim _{h \rightarrow 0} \frac{u(x+h)-u(x)}{h}=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$

$\frac{d}{d x}[f(x) g(x)] \neq f^{\prime}(x) g^{\prime}(x)$

e.g. $f(x)=g(x)=x$

then $f^{\prime} (x)=1= g^{\prime} (x)$

but $f(x) g(x)=x^2$

So, $\frac{d}{d x}[f(x) g(x)]=\frac{d}{d x}\left(x^2\right)=2 x \neq f^{\prime}(x) g'(x)$

e.g.

$f(x) =x^3 , f^{\prime}(x)= ?$

$f(x) =\frac{d}{d x}\left(x^2\right) \cdot x+x^2 \frac{d}{d x}(x)$

$=2 x \cdot x+x^2 \cdot 1$

$=2 x^2+x^2=3 x^2$

Calculate $\frac{d}{d x}\left(x^n\right)$, where $n \in \mathbb{N}$.

$f(x)=x^n$

For $h \neq 0, \frac{f(x+h)-f(x)}{h}$

$=\frac{(x+h)^n-x^n}{h}$

$=\frac{x^n+ \binom n1 x^{n-1} h+ \binom n2 x^{n-2} h^2+\cdots+h^n - x^n}{h}$

$=\frac{h[n x^{n-1}+ \binom n2 x^{n-2} h+\cdots+h^{n-1}]}{h}$

$= {n x^{n-1} }$ as h $\rightarrow$ 0

$\therefore \frac{d}{d x}\left[x^n\right] =n x^{n-1}$ for n $\in N$

$\text { eg. } \quad \frac{d}{d x}\left(x^2\right) =2 x^{2-1}=2 x$

$\frac{d}{d x}\left(x^3\right) =3 x^{3-1}=3 x^2$

$\frac{d}{d x}\left(x^4\right) =4 x^3 .$

Remark: The above formula for derivative of $x^n$ actually holds true for any $n \in \mathbb{R}$. This will be proved later.

$\frac{f(x+h)-f(x)}{h} =\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$

$=\frac{x-(x+h)}{h x(x+h)}=\frac{-h}{h x(x+h)}$

$=\frac{-1}{x(x+h)}, h \neq 0 .$

$\therefore \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\frac{-1}{x^2} \quad \forall x \neq 0 .$

$\frac{d}{d x}\left(\frac{1}{x}\right)=\frac{-1}{x^2}, x \neq 0$

Note that this agrees with same formula for

$\frac{d}{d x}\left(x^n\right)=n x^{n-1} \text {. }$

For $n=-1$, this gives $\frac{d}{d x}\left(x^{-1}\right)=(-1) x^{-1-1}=-{x}^{-2}=\frac{-1}{x^2} \text { }$

We can use product rule to calculate $\frac{d}{d x}\left(x^{-2}\right) or \frac{d}{d x}\left(x^{-3}\right)$, etc.

eg.

$\frac{d}{d x}\left(x^{-2}\right)=\frac{d}{d x}\left(\frac{1}{x} \cdot \frac{1}{x}\right)$

$=\left(\frac{-1}{x^2}\right) \cdot\frac{1}{x}+\frac{1}{x} \cdot\left(\frac{-1}{x^2}\right)$

$=\frac{-2}{x^3} .$

Suppose $f(x)$ is differentiable at $x=a$

and let $g(x)=\frac{1}{f(x)}$ & $f(a) \neq 0$

Then $g^{\prime}(a)=\frac{-f^{\prime}(a)}{(f(a))^2}$

Proof:

$\frac{g(a+h)-g(a)}{h}=\frac{\frac{1}{f(a+h)}-\frac{1}{f(a)}}{h}$

$=\frac{f(a)-f(a+h)}{h f(a) f(a+h)}$

${=-\underbrace{\left [\frac{f(a+h)-f(a)}{h}\right]}_{f^{\prime}(a)} \cdot \frac{1}{f(a) {f(a+h)}}}$

$\therefore g'(a) = \frac{-f'(a)}{(f(a))^2}$

(Quotient Rule)

$\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{[g(x)]^2}$

Pf:

$\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{d}{d x}\left[f(x) \cdot \frac{1}{g(x)}\right]$

(Using Product Rule )

$= f^{\prime}(x) \cdot \frac{1}{g(x)}+f(x) \frac{d}{d x}\left[\frac{1}{g(x)}\right]$

$=\frac{f^{\prime}(x)}{g(x)}+f(x)\left[\frac{-g^{\prime}(x)}{(g(x))^2}\right]$

$=\frac{f^{\prime}(x) g(x)-f(x) g^{\prime}(x)}{(g(x))^2}$

$\therefore \frac{d}{d x}[\sqrt{x}]=\frac{1}{2 \sqrt{x}} ,x>0$

Again note that, if we write $\sqrt{x}=x^{1 / 2}$

$\frac{d}{d x}(\sqrt{x})=\frac{1}{2 \sqrt{x}}=\frac{1}{2} \cdot x^{-1 / 2}=\frac{1}{2} x^{\frac{1}{2}-1} \text { }$

This also agrees with the formula

$\frac{d}{d x}\left[x^n\right]=n x^{n-1}$

Derivative of $f(x)=\sin (x)$

$\frac{f(x+h)-f(x)}{h}=\frac{\sin (x+h)-\sin x}{h}$

Recall:

$\sin C-\sin D=2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$

$\sin (x+h)-\sin x=2 \cos \frac{2 x+h}{2} \sin \frac{h}{2}$

$\therefore \frac{f(x+h)-f(x)}{h} =\frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \frac{h}{2}}{h}$

$=\cos \left(x+\frac{h}{2}\right) \cdot \frac{\sin (h/2)}{h / 2}$ $\left(\because \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\right)$

$\therefore \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}= cosx . 1$

$\therefore \frac{d}{d x}[\sin x]=\cos x$

eg. Derivative of $\cos x$.

$\frac{d}{d x}(\cos x) =\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}$

$=-\sin x$$\quad \text { [Exercise] }$

• $\therefore \frac{d}{d x}[\tan x]=\sec ^2 x$