# Derivatives of Products and Quotients. Find the derivative?

## $y = x {\left({x}^{2} - 2\right)}^{2}$

Jan 10, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left({x}^{2} - 2\right) \left(5 {x}^{2} - 2\right) , \mathmr{and} , 5 {x}^{4} - 12 {x}^{2} + 4$

#### Explanation:

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

We first differentiate using the Product Rule.

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

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

As regards, $\frac{d}{\mathrm{dx}} {\left({x}^{2} - 2\right)}^{2}$, we use the Chain Rule.

Let, $u = {\left({x}^{2} - 2\right)}^{2} = {t}^{2} , w h e r e , t = {x}^{2} - 2$

So, $u = {t}^{2} , t = \left({x}^{2} - 2\right)$

Thus, $u \text{ is a function of "t, and, t" of } x$

By the Chain Rule, then, $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

Now, u=t^2 :. (du)/dt=2t, &, t=x^2-2 :. dt/dx=2x

Utilising these, we have, $\frac{d}{\mathrm{dx}} {\left({x}^{2} - 2\right)}^{2} = \frac{\mathrm{du}}{\mathrm{dx}}$

$= \left(2 t\right) \left(2 x\right) = 4 t x = 4 x \left({x}^{2} - 2\right)$

Finally, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{2} \left({x}^{2} - 2\right) + {\left({x}^{2} - 2\right)}^{2} , i . e . ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left({x}^{2} - 2\right) \left\{4 {x}^{2} + \left({x}^{2} - 2\right)\right\} = \left({x}^{2} - 2\right) \left(5 {x}^{2} - 2\right) , \mathmr{and} ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4} - 12 {x}^{2} + 4$

Alternatively, without using the Product Rule and Chain Rule, we

can solve the Problem easily, as shown below :

$y = x {\left({x}^{2} - 2\right)}^{2} = x \left({x}^{4} - 4 {x}^{2} + 4\right) = {x}^{5} - 4 {x}^{3} + 4 x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4} - 4 \left(3 {x}^{2}\right) + 4 = 5 {x}^{4} - 12 {x}^{2} + 4 ,$ As Before!

Enjoy Maths.!