# How do you find the derivative of cos^2(x^2-2)?

Feb 18, 2016

#### Answer:

$\frac{d}{\mathrm{dx}} {\cos}^{2} \left({x}^{2} - 2\right) = 2 x \left(\sin \left(4 - 2 {x}^{2}\right)\right)$

#### Explanation:

First of all define: $f \left(x\right) = \cos \left({x}^{2} - 2\right)$
First use the chain rule to find the derivative of f(x):
$f ' \left(x\right) = - \sin \left({x}^{2} - 2\right) \left(2 x\right)$

Next split ${\cos}^{2} \left({x}^{2} - 2\right)$ into $\cos \left({x}^{2} - 2\right) \cos \left({x}^{2} - 2\right)$.
In other words: $f {\left(x\right)}^{2} = f \left(x\right) f \left(x\right)$

Now the product rule can be used to find the derivative of f(x)^2:
$f ' {\left(x\right)}^{2} = f ' \left(x\right) f \left(x\right) + f \left(x\right) f ' \left(x\right)$
$f ' {\left(x\right)}^{2} = \left(- \sin \left({x}^{2} - 2\right) \left(2 x\right)\right) \left(\cos \left({x}^{2} - 2\right)\right) + \left(\cos \left({x}^{2} - 2\right)\right) \left(- \sin \left({x}^{2} - 2\right) \left(2 x\right)\right)$

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

Using the trigonometric identity: $\cos \left(x\right) \sin \left(x\right) = \frac{1}{2} \sin \left(2 x\right)$

We finally gain:
f'(x)^2 = -4x (1/2(sin(2(x^2-2))
$f ' {\left(x\right)}^{2} = - 2 x \left(\sin \left(2 {x}^{2} - 4\right)\right)$

SIn(x) is an odd function, so it's true that:
sin(x) = -sin(-x)
Which means we can also write the answer as:
$f ' {\left(x\right)}^{2} = 2 x \left(\sin \left(4 - 2 {x}^{2}\right)\right)$
Which removes the minus sign from the front, making the expression tidier.