# What is the derivative of y=2(x^(3)-1)(3x^(2)+1)^4?

Aug 11, 2015

${y}^{'} = 6 x \left(3 {x}^{2} + 1\right) \cdot \left(11 {x}^{3} + x - 8\right)$

#### Explanation:

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

If you write $y$ as

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

then you can say that the derivative of $y$ is

$\frac{d}{\mathrm{dx}} \left(y\right) = 2 \cdot \left[\left[\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)\right] \cdot g \left(x\right) + f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)\right]$

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

${y}^{'} = 2 \left[\frac{d}{\mathrm{dx}} \left({x}^{3} - 1\right) \cdot {\left(3 {x}^{2} + 1\right)}^{4} + \left({x}^{3} - 1\right) \cdot \frac{d}{\mathrm{dx}} {\left(3 {x}^{2} + 1\right)}^{4}\right]$

To determine $\frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)$, use the chain rule for

• $g \left(u\right) = {u}^{4}$, with $u = 3 {x}^{2} + 1$

$\frac{d}{\mathrm{dx}} \left(g \left(u\right)\right) = \frac{d}{\mathrm{du}} \left({u}^{4}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left(g \left(u\right)\right) = 4 {u}^{3} \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 1\right)$

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

Your target derivative will thus be

${y}^{'} = 2 \cdot \left[3 {x}^{2} \cdot {\left(3 {x}^{2} + 1\right)}^{4} + \left({x}^{3} - 1\right) \cdot 24 x \cdot {\left(3 {x}^{2} + 1\right)}^{3}\right]$

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

${y}^{'} = 6 x \left(3 {x}^{2} + 1\right) \cdot \left(3 {x}^{3} + x + 8 {x}^{3} - 8\right)$

${y}^{'} = \textcolor{g r e e n}{6 x \left(3 {x}^{2} + 1\right) \cdot \left(11 {x}^{3} + x - 8\right)}$