# How do you differentiate 2x^4 * 6^(3x)?

Oct 8, 2015

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = {6}^{3 x} \cdot 2 {x}^{3} \left(\ln \left(6\right) \cdot 3 x + 4\right)$

#### Explanation:

You can either use the product / chain rule, or implicit differentiation with logarithms. Since you've posted this in Basic Differentiation Rules, I'll be using the chain rule.

Given that

$y = 2 {x}^{4} \cdot {6}^{3 x}$

We need to find the derivatives of each term to apply the product rule, so, for the first term we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 {x}^{3}$

For the second term we have

$y = {e}^{\ln \left(6\right) \cdot 3 x}$

Decomposing it, we have two functions

$y = f \left(u\right) = {e}^{u}$
$u = g \left(x\right) = \ln \left(6\right) \cdot 3 x$

The chain rule state that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Or,

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{\ln \left(6\right) \cdot 3 x} \cdot \ln \left(6\right) \cdot 3$

So, using the product rule we have

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = {6}^{3 x} \cdot \ln \left(6\right) \cdot 6 {x}^{4} + 8 {x}^{3} \cdot {6}^{3 x}$

Or

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = {6}^{3 x} \cdot 2 {x}^{3} \left(\ln \left(6\right) \cdot 3 x + 4\right)$

Doing implicit differentiation with logarithms,

$y = 2 {x}^{4} \cdot {6}^{3 x}$
$\ln \left(y\right) = \ln \left(2 {x}^{4}\right) + \ln \left({6}^{3 x}\right)$
$\ln \left(y\right) = \ln \left(2\right) + 4 \ln \left(x\right) + 3 x \ln \left(6\right)$
${y}^{'} / y = \frac{4}{x} + 3 \ln \left(6\right)$
${y}^{'} = \frac{4 \left(2 {x}^{4} \cdot {6}^{3 x}\right)}{x} + 3 \ln \left(6\right) \left(2 {x}^{4} \cdot {6}^{3 x}\right)$
${y}^{'} = 8 {x}^{3} \cdot {6}^{3 x} + 3 \ln \left(6\right) \left(2 {x}^{4} \cdot {6}^{3 x}\right)$
${y}^{'} = {6}^{3 x} \cdot 2 {x}^{3} \left(\ln \left(6\right) \cdot 3 x + 4\right)$