Question

Find the derivative of the following functions from first principle.
$\left(i\right)$ $x^3-27$
$\left(ii\right)$ $(x-1)(x-2)$
$\left(iii\right)$ $\frac{1}{x^2}$
$\left(iv\right)$ $\frac{x+1}{x-1}$

Answer 1

$\left(i\right)$ Let $f\left(x\right)=x^3-27$
We know that $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\left(\right(x+h{)}^3-27)-(x^3-27)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(x+h{)}^3-x^3-27+27}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(x+h{)}^3-x^3}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{x^3+h^3+3x^{2}h+3x{h}^2-x^3}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{h(h^2+3x^2+3xh)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}{h}^2+3x^2+3xh$

Putting $h=0$
$\Rightarrow f^{\prime }\left(x\right)=(0{)}^2+3x^2+3x(0)$
$\Rightarrow f^{\prime }\left(x\right)=0+3x^2+0$
$\Rightarrow f^{\prime }\left(x\right)=3x^2$
Hence, $f^{\prime }\left(x\right)=3x^2$

$\left(ii\right)$ Let $f\left(x\right)=(x-1)(x-2)$
$\Rightarrow f\left(x\right)=x(x-2)-1(x-2)$
$\Rightarrow f\left(x\right)=x^2-2x-x+2$
$\Rightarrow f\left(x\right)=x^2-3x+2$
We know that $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\left[\right(x+h{)}^2-3(x+h)+2]-(x^2-3x+2)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(x+h{)}^2-3x-3h+2-x^2+3x-2}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(x+h{)}^2-x^2-3h}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{x^2+h^2+2xh-x^2-3h}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{h(h+2x-3)}{h}$

putting $h=0$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}0+2x-3$
$\Rightarrow f^{\prime }\left(x\right)=0+2x-3$
$\Rightarrow f^{\prime }\left(x\right)=2x-3$
Hence,$f^{\prime }\left(x\right)=2x-3$

$\left(iii\right)$Let $f\left(x\right)=\frac{1}{x^2}$
We know that $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\frac{1}{(x+h{)}^2}-\frac{1}{x^2}}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\frac{x^2-(x+h{)}^2}{(x+h{)}^2{x}^2}}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{x^2-(x+h{)}^2}{h{x}^2(x+h{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(x-(x+h\left)\right)(x+(x+h\left)\right)}{h{x}^2(x+h{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(x-x-h)(x+x+h)}{h.x^2(x+h{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(-h)(2x+h)}{h{x}^2(x+h{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(-1)(2x+h)}{x^2(x+h{)}^2}$
Putting $h=0$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(-1)(2x+0)}{x^2(x+0{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(-1)2x}{x^2(x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)==\frac{-2x}{x^4}$
$\Rightarrow f^{\prime }\left(x\right)==\frac{-2}{x^3}$
Thus, $f^{\prime }\left(x\right)=\frac{-2}{x^3}$

$\left(iv\right)$ Let $f\left(x\right)=\frac{x+1}{x-1}$
We know that $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(\left[\frac{(x+h)+1}{(x+h)-1}\right]-\left[\frac{x+1}{x-1}\right])}{h}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}$
$={lim}_{h\to 0}\frac{\frac{(x-1)(x+h+1)-(x+1)(x+h-1)}{h(x+h-1)(x-1)}}{h}$
$={lim}_{h\to 0}\frac{(x-1)\left(\right(x+1)+h)-(x+1)\left(\right(x-1)+h))}{h(x+h-1)(x-1)}$
$={lim}_{h\to 0}\frac{(x-1)(x+1)+(x-1)h-(x+1)(x-1)-(x+1)h}{h(x+h-1)(x-1)}$
$={lim}_{h\to 0}\frac{(x^2-1)+xh-h-(x^2-1)-xh-h}{h(x+h-1)(x-1)}$
$={lim}_{h\to 0}\frac{-2h}{h(x+h-1)(x-1)}$
$={lim}_{h\to 0}\frac{-2}{(x+h-1)(x-1)}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{-2}{(x+h-1)(x-1)}$

putting $h=0$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-2}{(x+0-1)(x-1)}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-2}{(x-1)(x-1)}$
$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{-2}{(x-1{)}^2}$
Hence, $f^{\prime }\left(x\right)=\frac{-2}{(x-1{)}^2}$