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

1 Answer
Feb 18, 2016

Answer:

#d/dxcos^2(x^2-2) = 2x (sin(4-2x^2)) #

Explanation:

First of all define: #f(x) = cos(x^2-2)#
First use the chain rule to find the derivative of f(x):
#f'(x)= -sin(x^2-2) (2x)#

Next split #cos^2(x^2-2)# into #cos(x^2-2) cos(x^2-2)#.
In other words: #f(x)^2 = f(x)f(x)#

Now the product rule can be used to find the derivative of f(x)^2:
#f'(x)^2 = f'(x)f(x) + f(x)f'(x)#
#f'(x)^2 = (-sin(x^2-2) (2x))(cos(x^2-2)) + (cos(x^2-2))(-sin(x^2-2) (2x)) #

#f'(x)^2 = 2 (cos(x^2-2))(-sin(x^2-2) (2x)) #
#f'(x)^2 = -4x (cos(x^2-2)sin(x^2-2)) #

Using the trigonometric identity: #cos(x)sin(x) = 1/2 sin(2x)#

We finally gain:
#f'(x)^2 = -4x (1/2(sin(2(x^2-2)) #
#f'(x)^2 = -2x (sin(2x^2-4)) #

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'(x)^2 = 2x (sin(4-2x^2)) #
Which removes the minus sign from the front, making the expression tidier.