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

Answer 1

Samantha Adkison

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)#

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Answer 2

Elijah Abram

The derivative is

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

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Answer 3

Scarlett Allison

To find the derivative of ( \cos^2(x) ), you use the chain rule.

[ \frac{d}{dx}(\cos^2(x)) = 2 \cos(x) \cdot (-\sin(x)) = -2 \cos(x) \sin(x) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.