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

Solution:

Let $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\frac{\mathrm{b}}{\mathrm{x}}$

We need to find the derivative of $f(x)$ i.e. $f^{\prime}(x)$

We know that,

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

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

$f(x+h)=a(x+h)^{2}+\frac{b}{(x+h)}$

Putting values in (i), we get

$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\left[\mathrm{a}(\mathrm{x}+\mathrm{h})^{2}+\frac{\mathrm{b}}{(\mathrm{x}+\mathrm{h})}\right]-\left[\mathrm{ax}^{2}+\frac{\mathrm{b}}{\mathrm{x}}\right]}{\mathrm{h}}$

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

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

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

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

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

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

Taking ‘h’ common from both the numerator and denominator, we get

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

Putting h = 0, we get

$=a[(0)+2 x]-\frac{b}{x(x+0)}$

$=2 a x-\frac{b}{x^{2}}$

Hence,

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