How do you find the derivative of #cos(1-2x)^2#?

1 Answer
Jan 13, 2016

#4(1-2x)sin(1-2x^2#

Explanation:

Use the chain rule multiple times.

The first issue is the cosine function.

The chain rule states that #d/dx[cos(u)]=-sin(u)*u'#

#d/dx[cos(1-2x)^2]=-sin(1-2x)^2*d/dx[(1-2x)^2]#

The second issue is the squared term.

The chain rule states that #d/dx[u^2]=2u*u'#

#d/dx[(1-2x)^2]=2(1-2x)d/dx[1-2x]#

Since #d/dx[1-2x]=-2#,

#d/dx[cos(1-2x)^2]=-sin(1-2x)2*2(1-2x)(-2)#

#=4(1-2x)sin(1-2x)^2#