Question

Find the derivative of $f\left(x\right)=\frac{1}{x}$.

Answer 1

Given : $f\left(x\right)=\frac{1}{x}$.
We know that $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(\frac{1}{x+h}-\frac{1}{x})}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{x-(x+h)}{x(x+h)}}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x-x-h}{x(x+h)h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-h}{x(x+h)h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-1}{x(x+h)}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{x(x+0)}$
$\Rightarrow f^{\prime }\left(x\right)=-\frac{1}{x^2}$
Hence, $f^{\prime }\left(x\right)=-\frac{1}{x^2}$.