Question

Differentiate from first principle:

$\left(x\right)x^2+x+3$

Answer 1

Given:

$f\left(x\right)=x^2+x+3$

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:

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

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

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

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

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

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

Therefore, the derivative of $x^2+x+3$ is $2x+1.$