Question

Find the derivative of the following functions from first principle.

(i) $x^3$ - 27
(ii) (x-1) (x-2)

(iii) $\frac{1}{x^2}$
(iv) $\frac{x+1}{x-1}$

Answer 1

(i) Here f(x) = $x^3$ - 27

Then f(x+h) = $(x+h{)}^3$ - 27

We know that f' = $\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f\left(x\right)}{h}$

$\Rightarrow$ f' (x) = $\underset{h\to 0}{\text{lim}}\frac{(x+h{)}^3-27-x^3+27}{h}$

= $\underset{h\to 0}{\text{lim}}\frac{(x+h{)}^3-27-x^3+27}{h}$

= $\underset{h\to 0}{\text{lim}}\frac{h(h^2+3x^2+3xh)}{h}=3x^2$

(ii) Here f(x) = (x-1) (x-2) = $x^2$ - 3x+2

Then f(x+h) = (x+h-1) (x+h-2)

= $x^2+h^2$ + 2xh - 3x - 3h + 2

We know that f'(x) = $\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f\left(x\right)}{h}$

$\underset{h\to 0}{\text{lim}}\frac{h(h+2x-3)}{h}$ = 2x - 3

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

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

We know that f'(x) = $\underset{x\to 0}{\text{lim}}\frac{f(x+h)-f\left(x\right)}{h}$

$\Rightarrow$ f'(x) =$\underset{x\to 0}{\text{lim}}\frac{\frac{1}{(x+h{)}^2}-\frac{1}{x^2}}{h}$

=$\underset{h\to 0}{\text{lim}}\frac{x^2-(x+h{)}^2}{h{x}^2(x+h{)}^2}$

=$\underset{h\to 0}{\text{lim}}\frac{x^2-x^2-h^2-2xh}{h{x}^2(x+h{)}^2}$

=$\underset{h\to 0}{\text{lim}}\frac{h(-h-2x)}{h{x}^2(x+h{)}^2}$

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

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

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

We know that f'(x) = $\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f\left(x\right)}{h}$

$\Rightarrow$ f'(x) = $\underset{h\to 0}{\text{lim}}\frac{\frac{x+h+1}{x-h-1}-\frac{x+1}{x-1}}{h}$

= $\underset{h\to 0}{\text{lim}}\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{h(x+h-1)(x-1)}$

= $\underset{h\to 0}{\text{lim}}\frac{x^2-x+xh-h+x-1-x^2-xh+x-x-h+1}{h(x+h-1)(x-1)}$

= $\underset{h\to 0}{\text{lim}}\frac{-2h}{h(x+h-1)(x-1)}$

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