# Math Help - Trig Derivative Question

1. ## Trig Derivative Question

$y'(2\sin^{2}\theta) =$ ??

2. ## Re: Trig Derivative Question

Originally Posted by Jason76
$y'(2\sin^{2}\theta) =$ ??
What is y differentiated with respect to? Write it out as a dy / d? form, or we can't see what's being derived with respect to what.

3. ## Re: Trig Derivative Question

Maybe he meant the derivative of 2sin^2(Θ)? Then it's just 2*2sin(Θ)*cos(Θ) = 2sin(2Θ). Otherwise, I don't know...

4. ## Re: Trig Derivative Question

Your notation is very confusing. $y'(2 sin(2\theta))$ means the derivative of some unknown function, y(x), evaluated at $x= sin(2\theta)$. I believe what you mean is " $y(\theta)= 2 sin(2\theta)$. Find the derivative of y with respect to $\theta$". To do that, you need to know:
1) the derivative of 2x with respect to x.
2) the derivative of sin(x) with respect to x.
3) the derivative of $2\theta$ with respect to $\theta$ and
4) the chain rule.

Do you know those?

5. ## Re: Trig Derivative Question

$y = 2\sin^{2}\theta$

$y' = 2\(u)^{2} du$

$u = \sin \theta$

$du = \cos\theta$

$y' = 4\sin\theta\cos\theta$

6. ## Re: Trig Derivative Question

Originally Posted by Jason76
$y = 2\sin^{2}\theta$

$y' = 2\(u)^{2} du$

$u = \sin \theta$

$du = \cos\theta$

$y' = 4\sin\theta\cos\theta$
$du = \cos\theta$ as it stands is meaningless and wrong. So is $y' = 2\(u)^{2} du$ which is even worse.

You need to write $\frac{du}{d \theta} = \cos\theta$ and $y = 2 (u)^{2}$.

Please revise your notion of derivatives and exactly what the notation means. Many of your problems in understanding what's going on may stem from your continued inaccuracy with your notation.

Steer clear of the $y'$ notation unless it is completely clear what y is derived with respect to. Here it is not clear at all.