Question

Find the derivative of the following function from the first principle.

$\left(i\right)x^3-16$

$\left(ii\right)(x-1)(x-2)$

$\left(iii\right)\frac{1}{x^2}(x\ne 0)$

$\left(iv\right)\frac{x+1}{x-1}(x-1\ne 0)$

Answer 1

i) $f\left(x\right)=x^3-16$

$f^1\left(x\right)=\underset{h\to 0}{lim}\frac{[{(x+h)}^3-16]-(x^3-16)}{h}$

$=\underset{h\to 0}{lim}\frac{{3x}^{2}h-{3hx}^2}{h}$

$={3x}^2$

ii) $f\left(x\right)=x^2-3x+2$

$f^1\left(x\right)=\underset{h\to 0}{lim}\frac{[{(x^2+h)}^2-3(x+h)+2]-(x^2-3x+2)}{h}$

$=\underset{h\to 0}{lim}\frac{h^2+2xh-3h}{h}$

$=2x-3$

iii) $f\left(x\right)=1/x^2$

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

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

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

iv) $f\left(x\right)=(x+1)/(x-1)$

$f^1\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}$

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

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