# How do you differentiate f(x)= (2x^2-5)(4x^2+5)  using the product rule?

Dec 8, 2016

$f ' \left(x\right) = \left(4 x\right) \left(8 {x}^{2} - 5\right)$

#### Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$, or, $\left(u v\right) ' = \left(\mathrm{du}\right) v + u \left(\mathrm{dv}\right)$

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

So with $f \left(x\right) = \left(2 {x}^{2} - 5\right) \left(4 {x}^{2} + 5\right)$ we have;

$\left\{\begin{matrix}\text{Let "u = 2x^2-5 & => & (du)/dx =4x \\ "And } v = 4 {x}^{2} + 5 & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = 8 x\end{matrix}\right.$

$\setminus \setminus \setminus \setminus \setminus \frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$
$\therefore \frac{d}{\mathrm{dx}} \left(u v\right) = \left(2 {x}^{2} - 5\right) \left(8 x\right) + \left(4 x\right) \left(4 {x}^{2} + 5\right)$
$\therefore \frac{d}{\mathrm{dx}} \left(u v\right) = \left(4 x\right) \left\{2 \left(2 {x}^{2} - 5\right) + \left(4 {x}^{2} + 5\right)\right\}$
$\therefore \frac{d}{\mathrm{dx}} \left(u v\right) = \left(4 x\right) \left(4 {x}^{2} - 10 + 4 {x}^{2} + 5\right)$
$\therefore \frac{d}{\mathrm{dx}} \left(u v\right) = \left(4 x\right) \left(8 {x}^{2} - 5\right)$