- #1

- 28

- 0

r = sin(θ

^{2})cos(2θ)

answer in book

-2sin(θ

^{2})sin(2θ)+2θcos(2θ)cos(θ

^{2})

My attempt

tried using the product rule

(sin(θ

^{2}))(-sin(2θ)) + cos(2θ)cos(θ

^{2})

I got stuck right here

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- Thread starter TommG
- Start date

- #1

- 28

- 0

r = sin(θ

answer in book

-2sin(θ

My attempt

tried using the product rule

(sin(θ

I got stuck right here

- #2

adjacent

Gold Member

- 1,553

- 63

I think you will need chain rule too. Try doing it using chain rule + product rule

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

[tex]\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}[/tex]

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

Second you are differentiating [itex]sin(x^2)[/itex] with respect to [itex]\theta[/itex], not "[itex]\theta^2[/itex]" so you need to multiply by the derivative of [itex]\theta^2[/itex] with respect to [itex]\theta[/itex].

- #4

- 28

- 0

thank you for your help I have figured it out

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