Question

Differentiate from first principle:

$\left(v\right)\frac{x^2-1}{x}$

Answer 1

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

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

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{(x+h{)}^2-1}{x+h}-\frac{x^2-1}{x}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{x^2+2xh+h^2-1}{x+h}-\frac{x^2-1}{x}}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x^3+2x^{2}h+h^{2}x-x-x^3-x^{2}h+x+h}{xh(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x^{2}h+h^{2}x+h}{xh(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x^2+hx+1}{x(x+h)}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{x^2+0+1}{x(x+0)}$


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

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

Therefore, the derivative of $\frac{x^2-1}{x}\text{is}1+\frac{1}{x^2}.$