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# Differentiating inverse functions watch

1. how would you go about differentiating

x sin^-1 (x^2)

the sin^-1 is an inverse

im not rlly sure but i made sqrt(sin y) = x and sub it in, and differentiate as normal.

[sqrt(1-x^4) + x^2] / 2x
2. Do you know how to differentiate ? If not, working out how to do that on its own would be a good start. Then you can use the product rule and chain rule to complete the question.

Spoiler:
Show
3. In the spoiler I've shown how to differentiate .
I did this (and perhaps I should have just given hints - in fact, I'll split my spoiler up into more spoilers - you should try to do it yourself if you can) because it's useful to know, and if you're doing A level maths, then you may need to reproduce it in an exam. You should know how to find the derivatives of and too.

To find the derivative of , apply the chain rule.
Spoiler:
Show

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\cosy = \sqrt{1 - (sin^{2}y)

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\frac{d}{dx} (\sin^{-1}x) = \frac{1}{\sqrt(1 - x^{2})

Then you can just do what Nuodai said.

EDIT: It's so irritating that I spent quite a while typing out how to differentiate inverse sine in LaTeX, only to have it called potentially dangerous.
I hate LaTeX.

I wish it didn't look so pretty.

Edit: WTF IS ERROR 2?
AND WHAT IS ERROR 6?
GOD DAMN

Edit: Better. At least some of it works.

Edit: I give up. Here it is:
Spoiler:
Show

y = sin^-1x
x = siny
Spoiler:
Show

1 = cosy(dy/dx)
(dy/dx) = 1/cosy
Spoiler:
Show

sin^2y + cos^2y = 1
Spoiler:
Show

cosy = \sqrt(1 - sin^2y)
Spoiler:
Show

cosy = \sqrt(1 - x^2)
(d/dx) sin^-1x = 1\[\sqrt(1-x^2)]
4. (Original post by AnonyMatt)
I hate LaTeX.
You need to leave a space after commands like \sin and \cos.

You might want to justify this step further. At the moment it is unsatisfactory as . You have ignored the minus possibility.
5. (Original post by Kolya)
You need to leave a space after commands like \sin and \cos.

You might want to justify this step further. At the moment it is unsatisfactory as . You have ignored the minus possibility.
Thanks!

Hmm. Do you know, I've never actually noticed that the step is not justified. I think it's allowed to be written like that at A level.

It's obvious to see that it must be positive, but that's not really a justification, is it? (Is it?! )

Is it because
Unparseable or potentially dangerous latex formula. Error 6: Image was not produced or one of its dimensions is too small.
\cosy
* is positive for ?

EDIT: FML LATEX!!! * = cosy

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