differentiating trig Watch

em.d_4
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find dy/dx of siny=xcos2x

No idea where to start
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(Original post by emma201295)
find dy/dx of siny=xcos2x

No idea where to start
Do you know about implicit differentiation?

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em.d_4
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(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
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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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anosmianAcrimony
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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
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!
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james22
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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
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.
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em.d_4
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(Original post by majmuh24)
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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hahaha sorry differentiation the stress is clearly sending me stupid :P
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em.d_4
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(Original post by anosmianAcrimony)
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!
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?
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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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em.d_4
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could someone lay out steps of how to go about this as all the mathsy language is just confusing me I'm afraid
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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
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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em.d_4
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(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
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FireGarden
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(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.
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em.d_4
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Got it thanks everyone
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(Original post by emma201295)
where you've used product rule does cos2x not differentiate to -2sin2x? Sorry
Whoops sorry

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(Original post by emma201295)
Got it thanks everyone
What did you get?

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em.d_4
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(Original post by majmuh24)
What did you get?

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the answer in the text book which is cos2x-2xsin2x/cosy
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(Original post by emma201295)
the answer in the text book which is cos2x-2xsin2x/cosy
If you mean \dfrac{\cos (2x)-2x \sin (2x)}{\cos (y)} , that looks good to me

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em.d_4
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Any idea how to integrate 2/x -1/(x+1)^2 -2/x+1 ?
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(Original post by emma201295)
Any idea how to integrate 2/x -1/(x+1)^2 -2/x+1 ?
What do you mean? That equation is pretty ambiguous, any chance you could use
Latex or place brackets around the parts that you mean

Do you mean  \dfrac{2}{x} - \dfrac{1}{(x+1)^2} - \dfrac{2}{x+1}

I assume it's something to do with natural logarithms though
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