Finding the Derivative of sin(x^2)cos(2x)

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Need to find derivative

r = sin(θ2)cos(2θ)

answer in book

-2sin(θ2)sin(2θ)+2θcos(2θ)cos(θ2)My attempt

tried using the product rule

(sin(θ2))(-sin(2θ)) + cos(2θ)cos(θ2)

I got stuck right here
 
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I think you will need chain rule too. Try doing it using chain rule + product rule
 
adjacent it right- you need the chain rule.
\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}

First you are differentiating cos(2\theta) with respect to \theta, not "2\theta" so you need to multiply by the derivative of 2\theta with respect to \theta.

Second you are differentiating sin(x^2) with respect to \theta, not "\theta^2" so you need to multiply by the derivative of \theta^2 with respect to \theta.
 
thank you for your help I have figured it out
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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