# How do you differentiate f(x)=(x^2+1)(x^2-1) using the product rule?

Feb 27, 2016

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

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) \left({x}^{2} - 1\right)$

Applying product rule: ${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$

$f = {x}^{2} + 1 , g = {x}^{2} - 1$

$= \setminus \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) \left({x}^{2} - 1\right) + \setminus \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right) \left({x}^{2} + 1\right)$ ...(i)

$\setminus \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) = 2 x$

(Applying sum/difference rule : ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$
$= \setminus \frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \setminus \frac{d}{\mathrm{dx}} \left(1\right)$ $= 2 x + 0$ $= 2 x$)

Also,
$\frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)$= $2 x$

So,finally we have from (i),

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

Simplifying it,we get
$4 {x}^{3}$