Question

Differentiate from first principle:

$\left(iv\right)\frac{x^2+1}{x}$

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 $f\left(x\right)$ in 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 $\frac{x^2+1}{x}\text{is}1-\frac{1}{x^2}.$