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Mathematics Question on Derivatives

Find the derivative of the following functions from first principle.
(i) x3 − 27 (ii) ( x −1)( x- 2 )
(iii) 1x2\frac{1}{x^2} (iv) x+1x−1\frac{x+1}{x-1}

Answer

(i) Let f(x) = x3 - 27. Accordingly, from the first principle,
f'(x)= lim⁡h→0\lim_{h\rightarrow 0} f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} x3+h3+3x2h+3xh2−x3h\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}
= lim⁡h→0\lim_{h\rightarrow 0} h3+3x2h+3xh2h\frac{h^3+3x^2h+3xh^2}{h}
= lim⁡h→0\lim_{h\rightarrow 0}(h2+3x2+3xhh^2+3x^2+3xh)
=0+3x23x^2+0=3x23x^2.

(ii) Let f(x) = (x - 1) (x - 2). Accordingly, from the first principle,
f'(x) = \lim_{h\rightarrow 0}$$\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} (x+h−1)(x+h−2)−(x−1)(x−2)h\frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}
= \lim_{h\rightarrow 0}$$\frac{(hx+hx+h^2-2h-h)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 2hx+h2−3h2hx+h^2-3h
= lim⁡h→0\lim_{h\rightarrow 0} (2x+h-3)
=(2x+0-3)
=2x-3

(iii) Let f(x)=1x2\frac{1}{x^2} Accordingly, from the first principle,
f'(x)= lim⁡h→0\lim_{h\rightarrow 0} f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 1(x+h)2−1x2h\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [x2−(x+h)2x2(x+h)2\frac{x^2-(x+h)^2}{x^2(x+h)^2} ]
= lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [−h2−2xx2(x+h)2\frac{-h^2-2x}{x^2(x+h)^2} ]
= lim⁡h→0\lim_{h\rightarrow 0} [−h−2hxx2(x+h)2\frac{-h-2hx}{x^2(x+h)^2} ]
= 0−2xx2(x+0)2\frac{0-2x}{x^2(x+0)^2} =−2x3-\frac{2}{x^3}

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