# The derivatives of (sinx)^2 (cosx)^2 and (tanx)^2 - proofs of sin, cos and tan squared x.

In this hub you will learn how to find the derivate of sin²x, cos²x and tan²x.

Let’s start off with sin²x

The easiest way to do this is by the chain rule (use when you have a function of a function)

Let y = u² and u = sinx.

Now if you find the derivative of both of these you get dy/du = 2u and du/dx = cosx

So if you multiply both of these derivatives together you will get dy/dx.

dy/dx = dy/du × du/dx

= 2u × cosx.

Finally, substitute u = sinx into the above and simplify:

= 2 × sinx × cosx

=2sinxcosx.

Now 2sinxcosx is the same as sin2x:

So dy/dx = sin2x.

Let’s move on now and find the derivative of cos²x.

Again use the chain rule:

Let y = u² and u = cosx.

Now if you find the derivative of both of these you get dy/du = 2u and du/dx = -sinx

So if you multiply both of these derivatives together you will get dy/dx.

dy/dx = dy/du × du/dx

= 2u × -sinx.

Finally, substitute u = cosx into the above and simplify:

= 2 × cosx × -sinx

=-2sinxcosx or (-2sinx)

Finally let’s find the derivative of tan²x. Again, use the chain rule:

Let y = u² and u = tanx.

Now if you find the derivative of both of these you get dy/du = 2u and du/dx = sec²x (the derivative of tanx is sec²x)

So if you multiply both of these derivatives together you will get dy/dx.

dy/dx = dy/du × du/dx

= 2u × sec²x.

Finally, substitute u = cosx into the above and simplify:

= 2 × tanx × sec²x

=2tanxsec²x

**So the 3 derivative for sin²x, cos²x and tan²x are:**

**y = sin²x differentiates to dy/dx = sin2x **

**y = cos²x differentiates to dy/dx = –sin2x**

**y = tan²x differentiates to dy/dx = 2tanxsec²x**

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## Comments 1 comment

My son would understand exactly what you were talking about - I'd have to get my old math book out!