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

Solution:

Let

$f(x)=\frac{x^{2}+1}{x}$

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)

$f(x)=\frac{x^{2}+1}{x}$

$\mathrm{f}(\mathrm{x}+\mathrm{h})=\frac{(\mathrm{x}+\mathrm{h})^{2}+1}{\mathrm{x}+\mathrm{h}}=\frac{\mathrm{x}^{2}+\mathrm{h}^{2}+2 \mathrm{xh}+1}{\mathrm{x}+\mathrm{h}}$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{x^{2}+h^{2}+2 x h+1}{x+h}-\frac{x^{2}+1}{x}}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{\left(x^{2}+h^{2}+2 x h+1\right)(x)-\left(x^{2}+1\right)(x+h)}{(x+h)(x)}}{h}$

$=\lim _{h \rightarrow 0} \frac{x^{3}+x h^{2}+2 x^{2} h+x-\left[x^{3}+x^{2} h+x+h\right]}{h(x+h)(x)}$

$=\lim _{h \rightarrow 0} \frac{x^{3}+x h^{2}+2 x^{2} h+x-x^{3}-x^{2} h-x-h}{h(x+h)(x)}$

$=\lim _{h \rightarrow 0} \frac{x h^{2}+x^{2} h-h}{h(x+h)(x)}$

$=\lim _{h \rightarrow 0} \frac{x h+x^{2}-1}{(x+h)(x)}$

Putting h = 0, we get

$=\frac{x(0)+x^{2}-1}{(x+0)(x)}$

$=\frac{x^{2}-1}{(x)^{2}}$

Hence,

$f^{\prime}(x)=\frac{x^{2}-1}{x^{2}}$