Answer to Question #185853 in Calculus for Smile

Question

Differentiate with respect to x , x³-2x from the first principle

Answer 1

(i) f(x)=x

By first principle-

$f'(x)=lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}=lim_{h\rightarrow 0}\dfrac{x+h-x}{h}=lim_{h\rightarrow 0}\dfrac{h}{h}=1$

(ii) $f(x)=x^3-2x$

By first principle-

 $f'(x)=lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$

$=lim_{h\rightarrow 0}\dfrac{(x+h)^3-2(x+h)-(x^3-2x)}{h}$

$=lim_{h\rightarrow 0}\dfrac{(x^3+h^3+3x^2h+3xh^2-2x-2h-x^3-2x)}{h}$

$=lim_{h\rightarrow 0}\dfrac{(h^3+3x^2h+3xh^2-2h)}{h}$

$=lim_{h\rightarrow 0}\dfrac{h(h^2+3x^2+3xh-2)}{h}$

$=lim_{h\rightarrow 0}(h^2+3x^2+3xh-2)=3x^2-2$

Hence $f'(x)=3x^2-2$