Find the derivative of the following functions from first principle \( \dfrac{1}{x^2}\)

Asked by Pragya Singh | 1 year ago |  45

1 Answer

Solution :-

\( f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\)

\( \lim\limits_{h \to 0}\dfrac{\dfrac{1}{(x+h)^2}-\dfrac{1}{x^2}}{h}\)

\( \lim\limits_{h \to 0}\dfrac{1}{h}[\dfrac{x^2-(x+h)^2}{x^2(x+h)^2}]\)

\( \lim\limits_{h \to 0}\dfrac{1}{h}[\dfrac{x^2-x^2-2hx-h^2}{x^2(x+h)^2}]\)

\( \lim\limits_{h \to 0}\dfrac{1}{h}[\dfrac{-h^2-2hx}{x^2(x+h)^2}]\)

\( \lim\limits_{h \to 0}[\dfrac{-h^2-2hx}{x^2(x+h)^2}]\)

\( \dfrac{0-2x}{x^2(x+0)^2}=\dfrac{-2}{x^3}\)

Answered by Abhisek | 1 year ago

Related Questions