Question

Differentiate from first principle:

$\left(xii\right)x^3+4x^2+3x+2$

Answer 1

Givne:

$f\left(x\right)=x^3+4x^2+3x+2$

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:

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{(x+h{)}^3+4(x+h{)}^2+3(x+h)+2-(x^3+4x^2+3x+2)}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\left[\right(x+h{)}^3-x^3]+4\left[\right(x+h{)}^2-x^2]+3h}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\left[h\right\{(x+h{)}^2+x^2+x(x+h\left)\right\}]4h(2x+h)+3h}{h}$

$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\left[(x+h{)}^2+x^2+x(x+h)+4(2x+h)+3\right]$

$\Rightarrow f^{\prime }\left(x\right)=x^2+x^2+x^2+8x+3$

$\Rightarrow f^{\prime }\left(x\right)=3x^2+8x+3$

Hence, the derivative of $x^3+4x^2+3x+2$ is $3x^2+8x+3$