# How do you find the derivative of  [((3x-1)^4)((2x+1)^-3)] using the chain rule?

Jan 17, 2018

You going to have to use both the product and the chain rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(3 x - 1\right)}^{3}}{{\left(2 x + 1\right)}^{3}} \left[12 - 6 \frac{\left(3 x - 1\right)}{\left(2 x + 1\right)}\right]$

#### Explanation:

Product Rule:
$y = f \left(x\right) \times g \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) \times g \left(x\right) + f \left(x\right) \times g ' \left(x\right)$

Chain Rule:
$y = f \left(g \left(x\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right)$

y=[((3x−1)^4)((2x+1)^(−3))]

First we will break it into the front ((3x−1)^4) and back ((2x+1)^(−3)) sections and differentiate them individually.

$\textcolor{red}{f i r s t S e c t i o n}$
[d(3x−1)^4]/(dx)=4(3x-1)^3xx3=12(3x-1)^3

$\textcolor{red}{S e c o n d S e c t i o n}$

$\frac{d {\left(2 x + 1\right)}^{- 3}}{\mathrm{dx}} = - 3 {\left(2 x + 1\right)}^{- 4} \times 2 = - 6 {\left(2 x + 1\right)}^{-} 4$

Finally, using the product rule we combine these two sections to form the total derivative.

(dy)/(dx)=(12(3x-1)^3)((2x+1)^(-3))+((3x−1)^4)(-6(2x+1)^-4)

The easiest simplification to do is to move the negative sign out the front of the second section so it become a subtraction rather than an addition.

(dy)/(dx)=(12(3x-1)^3)((2x+1)^(-3))-((3x−1)^4)(6(2x+1)^-4)

You could then move the brackets with negative exponents into the denominator.

(dy)/(dx)=(12(3x-1)^3)/((2x+1)^(3))-(6(3x−1)^4)/((2x+1)^4)

Finally, you could factorise by ((3x-1)^3)/((2x+1)^(3)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(3 x - 1\right)}^{3}}{{\left(2 x + 1\right)}^{3}} \left[12 - 6 \frac{\left(3 x - 1\right)}{\left(2 x + 1\right)}\right]$

I hope that helps :)