# How do you differentiate (3x-4/x)^2 (1-x+7x^2)^4?

May 15, 2015

Chain rule and product rule time combined!

Let's work with your function $f \left(x\right) = {\left(3 x - 4 {x}^{-} 1\right)}^{2} {\left(1 - x + 7 {x}^{2}\right)}^{4}$.

First, we will rename:

$u = {\left(3 x - 4 {x}^{-} 1\right)}^{2}$ and $w = {\left(1 - x + 7 {x}^{2}\right)}^{4}$

Using the product rule, as follows,

$\frac{d \left(u w\right)}{\mathrm{dx}} = w \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) + u \left(\frac{\mathrm{dw}}{\mathrm{dx}}\right)$

${\left(1 - x + 7 {x}^{2}\right)}^{4} \cdot \textcolor{g r e e n}{\frac{d {\left(3 x - 4 {x}^{-} 1\right)}^{2}}{\mathrm{dx}}} + {\left(3 x - 4 {x}^{-} 1\right)}^{2} \cdot \textcolor{b l u e}{\frac{d {\left(1 - x + 7 {x}^{2}\right)}^{4}}{\mathrm{dx}}}$

Now, we will lead with the colored derivatives first and then return to the whole equation above.

To solve the green derivative, let's name $a = 3 x - 4 {x}^{-} 1$ and solve it as a chain rule, where

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

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

The same for the blue part, where $b = 1 - x + 7 {x}^{2}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} \left(- 1 + 14 x\right)$

Now we have the derivatives $\frac{\mathrm{du}}{\mathrm{dx}}$ and $\frac{\mathrm{dw}}{\mathrm{dx}}$, let's go back to the partillay colored long equation:

${\left(1 - x + 7 {x}^{2}\right)}^{4} \cdot 2 \left(3 x - 4 {x}^{-} 1\right) \cdot \left(3 + 4 {x}^{-} 2\right) + {\left(3 x - 4 {x}^{-} 1\right)}^{2} \cdot 4 {\left(1 - x + 7 {x}^{2}\right)}^{3} \cdot \left(- 1 + 14 x\right)$

You may want to expand it and eliminate the factoring (parenthesis), but that is not really necessary as the work and steps to achieve so would not necessarily mean simplification. The line above can be your answer.