How do you use the product Rule to find the derivative of y (x) = (x^2+1)(x^3+1)?

Jul 27, 2015

I found: $y ' \left(x\right) = 5 {x}^{4} + 3 {x}^{2} + 2 x$

Explanation:

The product rule allows you to derive a function formed by the product of two functions as $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$ to get:
$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$
In your case you have (in red are the derivatives):
$y ' \left(x\right) = \textcolor{red}{2 x} \left({x}^{3} + 1\right) + \left({x}^{2} + 1\right) \textcolor{red}{3 {x}^{2}} =$
$= 2 {x}^{4} + 2 x + 3 {x}^{4} + 3 {x}^{2} = 5 {x}^{4} + 3 {x}^{2} + 2 x$

Jul 27, 2015

You can always multiply it out if you don't want to use the product rule.

Product Rule:
$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = f \left(x\right) \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} + g \left(x\right) \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}$

$= \left({x}^{2} + 1\right) \left(3 {x}^{2}\right) + \left({x}^{3} + 1\right) \left(2 x\right)$

$= \left(3 {x}^{4} + 3 {x}^{2}\right) + \left(2 {x}^{4} + 2 x\right)$

$= \textcolor{b l u e}{5 {x}^{4} + 3 {x}^{2} + 2 x}$

You can also do this with the Power Rule:

$h \left(x\right) = \left({x}^{2} + 1\right) \left({x}^{3} + 1\right)$

$h \left(x\right) = {x}^{5} + {x}^{2} + {x}^{3} + 1$

$\frac{d}{\mathrm{dx}} \left[h \left(x\right)\right]$

$\implies 5 {x}^{4} + 2 x + 3 {x}^{2}$

$= \textcolor{b l u e}{5 {x}^{4} + 3 {x}^{2} + 2 x}$