What is the derivative of (1+4x)^5(3+x-x^2)^8?

Aug 11, 2015

${y}^{'} = 4 {\left(1 + 4 x\right)}^{4} \cdot {\left(3 + x - {x}^{2}\right)}^{7} \cdot \left(- 21 {x}^{2} + 9 x + 17\right)$

Explanation:

You can differentiate this function by using the chain rule and the product rule.

Notice that your function can be written as

$y = f \left(x\right) \cdot g \left(x\right)$

which means that its derivative can be determined using the product rule

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$

In your case, $f \left(x\right) = {\left(1 + 4 x\right)}^{5}$ and $g \left(x\right) = {\left(3 + x - {x}^{2}\right)}^{8}$.

Differentiate these functions separately by using the chain rule. For $f \left(x\right)$, you have

• $f \left(u\right) = {u}^{5}$, with $u = \left(1 + 4 x\right)$

This will get you

$\frac{d}{\mathrm{dx}} \left({u}^{5}\right) = \frac{d}{\mathrm{du}} \left({u}^{5}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left({u}^{5}\right) = 5 {u}^{4} \cdot \frac{d}{\mathrm{dx}} \left(1 + 4 x\right)$

$\frac{d}{\mathrm{dx}} {\left(1 + 4 x\right)}^{5} = 5 \cdot {\left(1 + 4 x\right)}^{4} \cdot 4$

For #g(x) you have

• $g \left({u}_{1}\right) = {u}_{1}^{8}$, with ${u}_{1} = 3 + x - {x}^{2}$

This will get you

$\frac{d}{\mathrm{dx}} \left({u}_{1}^{8}\right) = \frac{d}{{\mathrm{du}}_{1}} \left({u}_{1}^{8}\right) \cdot \frac{d}{\mathrm{dx}} \left({u}_{1}\right)$

$\frac{d}{\mathrm{dx}} \left({u}_{1}^{8}\right) = 8 {u}_{1}^{7} \cdot \frac{d}{\mathrm{dx}} \left(3 + x - {x}^{2}\right)$

$\frac{d}{\mathrm{dx}} {\left(3 + x - {x}^{2}\right)}^{8} = 8 {\left(3 + x - {x}^{2}\right)}^{7} \cdot \left(1 - 2 x\right)$

Take these derivatives to your target calculation to get

${y}^{'} = 20 \cdot {\left(1 + 4 x\right)}^{4} \cdot {\left(3 + x - {x}^{2}\right)}^{8} + {\left(1 + 4 x\right)}^{5} \cdot 8 \cdot \left(1 - 2 x\right) \cdot {\left(3 + x - {x}^{2}\right)}^{7}$

${y}^{'} = 4 {\left(1 + 4 x\right)}^{4} \cdot {\left(3 + x - {x}^{2}\right)}^{7} \cdot \left[5 \cdot \left(3 + x - {x}^{2}\right) + 2 \left(1 - 2 x\right) \left(1 + 4 x\right)\right]$

This is equivalent to

${y}^{'} = 4 {\left(1 + 4 x\right)}^{4} \cdot {\left(3 + x - {x}^{2}\right)}^{7} \cdot \left[15 + 5 x + 5 {x}^{2} + 2 \left(1 + 2 x - 8 {x}^{2}\right)\right]$

${y}^{'} = 4 {\left(1 + 4 x\right)}^{4} \cdot {\left(3 + x - {x}^{2}\right)}^{7} \cdot \left(15 + 5 x - 5 {x}^{2} + 2 + 4 x - 16 {x}^{2}\right)$

${y}^{'} = \textcolor{g r e e n}{4 {\left(1 + 4 x\right)}^{4} \cdot {\left(3 + x - {x}^{2}\right)}^{7} \cdot \left(- 21 {x}^{2} + 9 x + 17\right)}$