Question

Find from first principles the differential coefficient of $2x^2+3x$.

Answer 1

We have,

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

Put $x=x+h$

$g(x+h)=2(x+h{)}^2+3(x+h)$

We know that

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

Therefore,

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

$g^{\prime }\left(x\right)={lim}_{h\to 0}\frac{2h^2+4xh+3h}{h}$

$g^{\prime }\left(x\right)={lim}_{h\to 0}2h+4x+3$

$g^{\prime }\left(x\right)=4x+3$

Hence, this is the answer.