Question

Find the derivate of the following functions from first principle
(i) -x
(ii) (-x)^-1
​

Answer 1

$\huge\sf\red{\underline{\underline{Question}}}\::$

$\sf\green{Find \ the \ derivate \ of \ the \ following }$

$\sf\green{functions \ from \ first \ principle }$

(i) $\sf\gray{-x}$

(ii) $\sf\gray{(-x)^{-1}}$

$\huge\sf\pink{\underline{\underline{Solution}}}\::$

$\sf\purple{ (i) \ Let \ f(x) = -x. \ According, \ f(x + h) \ = \ -(x + h)} \\\\ \star\:\:\bf\underline\orange{ By \ first \ principle,}$

$\sf\blue{f'(x)\:=\:\lim_{h \to 0}\:\dfrac{f(x+h)-f(x)}{h}}$

⠀⠀⠀$\sf\green{=\:\lim_{h \to 0}\:\dfrac{-(x+h)-(-x)}{h}}$

⠀⠀⠀$\sf\blue{=\:\lim_{h \to 0}\:\dfrac{-x-h-x}{h}}$

⠀⠀⠀$\sf\green{=\:\lim_{h \to 0}\:\dfrac{-h}{h}}$

⠀⠀⠀$\sf\blue{=\:\lim_{h \to 0}\:(-1)}$

⠀⠀⠀$\sf\red{=\:-1}$

$\sf\purple{ (ii) \ Let \ f(x) = (-x)^{-1} = \dfrac{1}{-x} = \dfrac{-1}{x}. \ According, \ f(x+h) = \dfrac{1}{(x+h)}} \\\\ \star\:\:\bf \underline\orange{By \ first \ principle,}$

$\sf\blue{f'(x)\:=\:\lim_{h \to 0}\:\dfrac{f(x+h)-f(x)}{h}}$

⠀⠀⠀$\sf\green{=\:\lim_{h \to 0}\:\dfrac{1}{h}\Bigg[ \dfrac{-1}{x+h}-\bigg( \dfrac{-1}{x}\bigg) \Bigg]}$

⠀⠀⠀$\sf\blue{=\:\lim_{h \to 0}\:\dfrac{1}{h}\Bigg[ \dfrac{-1}{x+h}+\dfrac{1}{x}\Bigg]}$

⠀⠀⠀$\sf\green{=\:\lim_{h \to 0}\:\dfrac{1}{h}\Bigg[ \dfrac{-x+(x+h)}{x(x+h)}\Bigg] }$

⠀⠀⠀$\sf\blue{=\:\lim_{h \to 0}\:\dfrac{1}{h}\Bigg[ \dfrac{h}{x(x+h)} \Bigg] }$

⠀⠀⠀$\sf\green{=\:\lim_{h \to 0}\:\dfrac{1}{x(x+h)}}$

⠀⠀⠀$\sf\blue{=\:\dfrac{1}{x.x}}$

⠀⠀⠀$\sf\red{=\:\dfrac{1}{x^2}}$