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

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Answer 1 · 2016-02-05T12:47:16

The derivative is

#(d(cos^2x))/dx=2cosx*(d(cosx))/dx=-2cosx*sinx=-sin2x#

Answer 2 · 2016-02-05T12:48:56

Remember that #cos^2x = (cosx)^2#.

Now use #d/dx (u^2) = 2u (du)/dx#.

In this case #u=cosx#, so that #(du)/dx = -sinx#

#d/dx(cos^2x) = 2(cosx)(-sinx) = -2sinxcosx#

Depending on how well you know trigonometry, you may or may not recognize #2sinxcosx# as equal to #sin2x#.

We can also write:

#d/dx(cos^2x) = -sin(2x)#