Find the derivation of each of the following from the first principle:

Solution:

Let $f(x)=\frac{1}{x^{3}}$

We need to find the derivative of f(x) i.e. f’(x)

We know that,

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

$f(x)=\frac{1}{x^{3}}$

$\mathrm{f}(\mathrm{x}+\mathrm{h})=\frac{1}{(\mathrm{x}+\mathrm{h})^{3}}$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)^{3}}-\frac{1}{x^{3}}}{h}$

$=\lim _{h \rightarrow 0} \frac{(x+h)^{-3}-x^{-3}}{(x+h)-x}$

[Add and subtract x in denominator]

$=\lim _{z \rightarrow x} \frac{z^{-3}-x^{-3}}{z-x}$ where $z=x+h$ and $z \rightarrow x$ as $h \rightarrow 0$

$=(-3) x^{-3-1}\left[\because \lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]$

$=-3 x^{-4}$

$=-\frac{3}{x^{4}}$

Hence,

$f(x)=-\frac{3}{x^{4}}$