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.
my answer is:
[sqrt(1-x^4) + x^2] / 2x
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Differentiating inverse functions watch
- Thread Starter
- 23-01-2010 21:38
- 23-01-2010 21:42
- 24-01-2010 00:31
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.
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?
Edit: Better. At least some of it works.
Edit: I give up. Here it is:
y = sin^-1x
x = siny
1 = cosy(dy/dx)
(dy/dx) = 1/cosy
sin^2y + cos^2y = 1
cosy = \sqrt(1 - sin^2y)
cosy = \sqrt(1 - x^2)
(d/dx) sin^-1x = 1\[\sqrt(1-x^2)]
Last edited by AnonyMatt; 24-01-2010 at 00:45.
- 24-01-2010 00:47
- 24-01-2010 02:49
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 * is positive for ?
EDIT: FML LATEX!!! * = cosy