Question

Find the derivative of the following functions from first principle. 
(i) x³ − 27 (ii) ( x −1)( x- 2 )
(iii) \(\frac{1}{x^2}\) (iv) \(\frac{x+1}{x-1}\)

Answer 1

(i) Let f(x) = x³ - 27. Accordingly, from the first principle,
f'(x)= \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{h^3+3x^2h+3xh^2}{h}\)
= \(\lim_{h\rightarrow 0}\)(\(h^2+3x^2+3xh\))
=0+\(3x^2\)+0=\(3x^2\).

(ii) Let f(x) = (x - 1) (x - 2). Accordingly, from the first principle,
f'(x) = \(\lim_{h\rightarrow 0}\)\(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}\)
= \(\lim_{h\rightarrow 0}\)\(\frac{(hx+hx+h^2-2h-h)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(2hx+h^2-3h\)
= \(\lim_{h\rightarrow 0}\) (2x+h-3)
=(2x+0-3) 
=2x-3

(iii) Let f(x)=\(\frac{1}{x^2}\) Accordingly, from the first principle,
f'(x)= \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [\(\frac{x^2-(x+h)^2}{x^2(x+h)^2}\) ]
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [\(\frac{-h^2-2x}{x^2(x+h)^2}\) ]
= \(\lim_{h\rightarrow 0}\) [\(\frac{-h-2hx}{x^2(x+h)^2}\) ]
= \(\frac{0-2x}{x^2(x+0)^2}\) =\(-\frac{2}{x^3}\)

(iv) Let f(x)= \(\frac{x+1}{x-1}\). Accordingly, from the first principle,
f'(x)= \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
=\(\lim_{h\rightarrow 0}\)\(\frac{(\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1})}{h}\)
= \(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\) [\(\frac{-2h}{(x-1)(x+h-1)}\)]
= \(\lim_{h\rightarrow 0}\) [\(\frac{-2}{(x-1)(x+h-1)}\)]
= \(\frac{-2}{(x-1)(x-1)}\)= \(\frac{-2}{(x-1)^2}\)