differentiating trig

em.d_4

find dy/dx of siny=xcos2x

No idea where to start :frown:

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Reply 1

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Original post by emma201295

find dy/dx of siny=xcos2x

No idea where to start :frown:

Do you know about implicit differentiation?

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Reply 2

em.d_4

Original post by majmuh24

Do you know about implicit differentiation?

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Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

Reply 3

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Original post by emma201295

Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

It allows you to differentiate functions of y with respect to x by defining the derivative of y WRT x as dy/dx and then using the chain rule on these functions of y.

Are you talking about integration or differentiation?

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Reply 4

anosmianAcrimony

Original post by emma201295

Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

You have sort of already done implicit differentiation; you do it whenever you normally differentiate something. Basically, wherever you would differentiate something containing a y, differentiate it as if it contains an x instead, but then multiply it by dy/dx. A normal equation where the subject is y gives a differentiated equation with dy/dx as the subject. In this case, you will then need to rearrange to find dy/dx in terms of both x's and y's.

I hope that makes some sense!

Reply 5

james22

Original post by emma201295

Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

Those are integration things, this is differentiation. If you haven't done implicit differentiation then this question is much more difficult, I would wait untill you cover it in class.

Reply 6

em.d_4

Original post by majmuh24

hahaha sorry differentiation the stress is clearly sending me stupid :P

Reply 7

em.d_4

Original post by anosmianAcrimony

Sorry none at all, could you lay it out as steps.
I man surely the equation needs to be layed out as y=.. or x=...etc how can you dfferetiate when it is split as it is?

Reply 8

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Original post by emma201295

hahaha sorry differentiation the stress is clearly sending me stupid :P

So do you understand what implicit differentiation is then?

Basically, the derivative of y WRT x is dy/dx and you treat all functions of y as a function of a function and use the chain rule on them to get a function of dy/dx, which you can then rearrange to get the derivative.

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Reply 9

em.d_4

could someone lay out steps of how to go about this as all the mathsy language is just confusing me I'm afraid :frown:

Reply 10

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Original post by emma201295

could someone lay out steps of how to go about this as all the mathsy language is just confusing me I'm afraid :frown:

I'll start it off.

Using implicit differentiation, you can get

\dfrac{dy}{dx} (sin(y)=x \cos(2x)) \\ \\ = (cos(y) \dfrac{dy}{dx} = \cos(2x) - 2x \sin(2x))

by using the product rule on the RHS.

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(edited 10 years ago)

Reply 11

em.d_4

Original post by majmuh24

I'll start it off.

Using implicit differentiation, you can get

\dfrac{dy}{dx} (sin(y)=x \cos(2x)) \\ \\ = (cos(y) \dfrac{dy}{dx} = \cos(2x) + 2x \cos(2x))

by using the product rule on the RHS.

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where you've used product rule does cos2x not differentiate to -2sin2x? Sorry

Reply 12

FireGarden

Original post by emma201295

Sorry none at all, could you lay it out as steps.
I man surely the equation needs to be layed out as y=.. or x=...etc how can you dfferetiate when it is split as it is?

Implicit differentiation lets you deal with the weird cases like in your OP. I don't like calling it implicit differentiation though. It's just an application of the chain rule.

So suppose I have

y^2=\cos(x)

and I need to differentiate it. Well then:

\dfrac{d}{dx}y^2=\dfrac{d}{dx} \cos(x)

.

Take

u(x)=y(x)^2

. Then we use the chain rule and get

\dfrac{du}{dx}=\dfrac{du}{dy} \dfrac{dy}{dx} = 2y\dfrac{dy}{dx}

.

Since

u(x)=y(x)^2

, we have

\dfrac{d}{dx}y^2=2y\dfrac{dy}{dx}

.

So going back, we find the derivative is

2y\dfrac{dy}{dx}=-\sin(x)

.

Of course that's not explicitly written, but you can rearrange.. You can't do any better than having y's around either. Hopefully you now see how to apply this to your problem.