Question

Differentiate from first principle:

$\left(vi\right)(-x{)}^{-1}$

Answer 1

Given:

$f\left(x\right)=(-x{)}^{-1}$

The derivative of a function $f\left(x\right)$ is defined as:

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

Putting $f\left(x\right)$ in the above expression, we get:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{1}{-(x+h)}-\frac{1}{-x}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{-1}{(x+h)}+\frac{1}{x}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{-x+x+h}{hx(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{h}{hx(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{1}{x(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x^2}$