If Y = x³ + 2x² + 7x + 8 then $\frac{dy}{dx}$ will be 

$\frac{dy}{dx}=3x^2+4x+7$      we  know that     $\frac{d\left(k x^n\right)}{dx} =k n x^{n-1 }, \frac{d\left(kx\right)}{dx} =k and \frac{d\left(k\right)}{dx} =0 \sin ce k is a cons\tan t.$

given Y=$x^3+2 x^{2 }+7 x+8$

$$\begin{gathered} \frac{d\left(x^3\right)}{dx}=3 x^{3-1}=3 x^2 \\ \frac{d(2 X^2)}{dx} =2\times 2x^{2-1}=4x \\ \frac{d\left(7x\right)}{dx} =7 \\ \frac{d\left(8\right)}{dx}=0 \\ \therefore \frac{dY}{dx} =3 x^2+4x+7 \end{gathered}$$