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Trig differentiation help c4

Not sure on 4. Any Trig identities used?

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Original post by The_Blade

Not sure on 4. Any Trig identities used?

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I would just expand the bracket and then apply the product rule
so

y=2sinxcos2x+sinx

Differentiate that with respect to x and set

\dfrac{dy}{dx}=0

to find the x-coordinate of the turning point(s)

OP

Original post by Robbie242

I would just expand the bracket and then apply the product rule
so

y=2sinxcos2x+sinx

Differentiate that with respect to x and set

\dfrac{dy}{dx}=0

to find the x-coordinate of the turning point(s)

It's what I tried to do but still failed

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Original post by The_Blade

It's what I tried to do but still failed

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\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}+\dfrac{d}{dx}(sinx)

where

u=2sinx, v=cos2x

\Rightarrow \dfrac{du}{dx}=2cosx, \dfrac{dv}{dx}=-2sin2x

plug all this into the formula above and you should get your desired answer.

OP

Original post by Robbie242

\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}+\dfrac{d}{dx}(sinx)

where

u=2sinx, v=cos2x

\Rightarrow \dfrac{du}{dx}=2cosx, \dfrac{dv}{dx}=-2sin2x

plug all this into the formula above and you should get your desired answer.

I don't think I did it right

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What are you getting for the derivative?
Once you have it, equate it to zero and solve. My recommendation would be to use the correct identities to try and form a cubic in cos(x) (don't worry, it factorises easily) and then use that to find the necesary values of x.

Original post by The_Blade

I don't think I did it right

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Ahh I see the mistake, you shouldn't have squared terms, do you see that

cos2xcosx

does not equal

cos^{2}(2x)

the multiple of x's are different

just to help you should have:

\dfrac{dy}{dx} =2sinx(-2sin2x)+2cosx(cos2x)+cosx[br]=-4sinxsin2x+2cosxcos2x+cosx

OP

Original post by Robbie242

Ahh I see the mistake, you shouldn't have squared terms, do you see that

cos2xcosx

does not equal

cos^{2}(2x)

the multiple of x's are different

just to help you should have:

\dfrac{dy}{dx} =2sinx(-2sin2x)+2cosx(cos2x)+cosx[br]=-4sinxsin2x+2cosxcos2x+cosx

To get it equal to 0 would I have to use cos2x identity and sin2x?

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Original post by The_Blade

To get it equal to 0 would I have to use cos2x identity and sin2x?

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Well the gradient is already 0 so it would be 0 by default, but you should be able to make use of those identities and

sin^{2}x+cos^{2}x=1

to get some sort of quadratic or even cubic and jared has stated

OP

Original post by Robbie242

Well the gradient is already 0 so it would be 0 by default, but you should be able to make use of those identities and

sin^{2}x+cos^{2}x=1

to get some sort of quadratic or even cubic and jared has stated

Not sure which cos2x identity to use

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Original post by The_Blade

Not sure which cos2x identity to use

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use

cos2x=2cos^{2}x-1

and use

sin^{2}x=1-cos^{2}x

after using

sin2x=2sinxcosx

OP

Original post by Robbie242

use

cos2x=2cos^{2}x-1

and use

sin^{2}x=1-cos^{2}x

after using

sin2x=2sinxcosx

Is this right so far?

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Original post by The_Blade

Is this right so far?

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looks correct to me :yy:

:yy:

Original post by The_Blade

Is this right so far?

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It's close but you've made a slight error towards the end.

-8(1-\cos^{2}(x))\cos(x)

should go to

-8\cos(x)+8\cos^{3}(x)

rather than

-8+8\cos^{3}(x)

.

Fix that and it should work.

OP

Original post by Jarred

It's close but you've made a slight error towards the end.

-8(1-\cos^{2}(x))\cos(x)

should go to

-8\cos(x)+8\cos^{3}(x)

rather than

-8+8\cos^{3}(x)

.

Fix that and it should work.

Fixed but not sure how to solve last part

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Original post by The_Blade

Fixed but not sure how to solve last part

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Take cos(x) out as a factor, and you'll have something that you can solve.

OP

Thank you very much jarred and robbie

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