# How do you find the derivative of y= x^3*2^x?

May 6, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} {2}^{x} \left(3 + x \ln 2\right)$

#### Explanation:

We have;

$y = {x}^{3} {2}^{x}$

We have a variable exponent so we must use the known result for exponents:

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x} \implies \frac{d}{\mathrm{dx}} {a}^{x} = \ln \cdot {a}^{x}$

So we can apply the product rule to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left({x}^{3}\right) \left(\frac{d}{\mathrm{dx}} {2}^{x}\right) + \left(\frac{d}{\mathrm{dx}} {x}^{3}\right) \left({2}^{x}\right)$
$\text{ } = \left({x}^{3}\right) \left(\ln 2 \cdot {2}^{x}\right) + \left(3 {x}^{2}\right) \left({2}^{x}\right)$
$\text{ } = \ln 2 \cdot {x}^{3} \cdot {2}^{x} + 3 {x}^{2} \cdot {2}^{x}$
$\text{ } = {x}^{2} {2}^{x} \left(3 + x \ln 2\right)$