derivative of tan^-1(y)

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absolutelysprout
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#1
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#1
hi im currently trying to differentiate tan-1y + xy = 0

xy i have no problem with but im not really sure on how to approach tan-1y
i googled how to differentiate tan-1x and it came up w 1 / x2+1

idk if im right in assuming that differentiating tan-1y would give a different answer- i did:
x = tan-1y
tanx = y
sec2x = 1 * dy/dx
sec2x = dy/dx
tan2x+1 = dy/dx
dy/dx = y2+1

any pointers??
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Notnek
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#2
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#2
(Original post by entertainmyfaith)
hi im currently trying to differentiate tan-1y + xy = 0

xy i have no problem with but im not really sure on how to approach tan-1y
i googled how to differentiate tan-1x and it came up w 1 / x2+1

idk if im right in assuming that differentiating tan-1y would give a different answer- i did:
x = tan-1y
tanx = y
sec2x = 1 * dy/dx
sec2x = dy/dx
tan2x+1 = dy/dx
dy/dx = y2+1

any pointers??
Are you trying to differentiate tan-1y + xy = 0 with respect to x or y?
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absolutelysprout
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#3
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#3
(Original post by Sir Cumference)
Are you trying to differentiate tan-1y + xy = 0 with respect to x or y?
well the textbook says to find dy/dx of the expression in terms of x and y if that helps?
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RDKGames
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#4
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(Original post by entertainmyfaith)
i googled how to differentiate tan-1x and it came up w 1 / x2+1
So your derivative for \arctan y wrt x is going to be \dfrac{1}{y^2 + 1} with a factor of  \dfrac{dy}{dx} by the side. This is just implicit differentiation.
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Notnek
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(Original post by entertainmyfaith)
well the textbook says to find dy/dx of the expression in terms of x and y if that helps?
That means you should differentiate wrt x (you could also do y but there will be an extra step). I recommend using different variables to find the general derivative of tan^-1 so you don't get confused:

s=\tan^{-1} t

Your aim is to find \frac{ds}{dt}. If you follow the same process as you did before then you'll end up with

\frac{dt}{ds} = t^2+1

So what's \frac{ds}{dt}?
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absolutelysprout
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(Original post by Sir Cumference)
That means you should differentiate wrt x (you could also do y but there will be an extra step). I recommend using different variables to find the general derivative of tan^-1 so you don't get confused:

s=\tan^{-1} t

Your aim is to find \frac{ds}{dt}. If you follow the same process as you did before then you'll end up with

\frac{dt}{ds} = t^2+1

So what's \frac{ds}{dt}?
1 / t2+1 ?
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absolutelysprout
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#7
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(Original post by RDKGames)
So your derivative for \arctan y wrt x is going to be \dfrac{1}{y^2 + 1} with a factor of  \dfrac{dy}{dx} by the side. This is just implicit differentiation.
thank you :awesome: we weren't really taught about differentiating inverse trig functions so had no idea what to do first
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Notnek
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#8
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#8
(Original post by entertainmyfaith)
1 / t2+1 ?
Yes that’s the derivative of tan^(-1)t.

So going back to your original question, can you find the derivative with respect to x now you know what the general derivative of tan inverse is? It’s just like differentiating sin(y) with respect to x.
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absolutelysprout
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#9
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#9
(Original post by Sir Cumference)
Yes that’s the derivative of tan^(-1)t.

So going back to your original question, can you find the derivative with respect to x now you know what the general derivative of tan inverse is? It’s just like differentiating sin(y) with respect to x.
wrt x; does that mean you put the dy/dx after?
1 / y2+1 * dy/dx?
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Notnek
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(Original post by entertainmyfaith)
wrt x; does that mean you put the dy/dx after?
1 / y2+1 * dy/dx?
Yep
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RDKGames
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(Original post by entertainmyfaith)
wrt x; does that mean you put the dy/dx after?
1 / y2+1 * dy/dx?
Because chain rule; \dfrac{d}{dx} \arctan(y) = \dfrac{dy}{dx} \cdot \dfrac{d}{dy} \arctan y
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absolutelysprout
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#12
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(Original post by Sir Cumference)
Yep
thank you i mostly understand it now!! i presume these type of questions could appear in the paper then?
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Notnek
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(Original post by entertainmyfaith)
thank you i mostly understand it now!! i presume these type of questions could appear in the paper then?
Well that's an interesting one and it's hard to know for sure whether differentiation of inverse trig functions could be in the exam. See here for a discussion about this:

https://www.thestudentroom.co.uk/sho....php?t=5817618
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Notnek
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(Original post by entertainmyfaith)
thank you i mostly understand it now!! i presume these type of questions could appear in the paper then?
I should have asked what you meant by "these type of questions". If you mean implicit differentiation questions then yes they could definitely come up.

I've also assumed you're doing Edexcel but I should have asked.
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absolutelysprout
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#15
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(Original post by Sir Cumference)
Well that's an interesting one and it's hard to know for sure whether differentiation of inverse trig functions could be in the exam. See here for a discussion about this:

https://www.thestudentroom.co.uk/sho....php?t=5817618
(Original post by Sir Cumference)
I should have asked what you meant by "these type of questions". If you mean implicit differentiation questions then yes they could definitely come up.

I've also assumed you're doing Edexcel but I should have asked.
yeah i meant differentiation of inverse trig functions
i'm doing aqa :ninja:
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Notnek
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#16
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#16
(Original post by entertainmyfaith)
yeah i meant differentiation of inverse trig functions
i'm doing aqa :ninja:
If it's in the AQA textbook then it could be in the exam.
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