Inverse : Help (1 Viewer)

Lindurr

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Hi I keep doing this question and getting it wrong, dunno if the answer is wrong

Differentiate inverse sin (1/x)

Thanks
 

Vidhya

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is the answer



edit: oops I maybe wrong
That was what I got, hope it's right.

I used the basic proof for integration of sine (x), but subbed x with 1/x. That is, to find dy/dx, find dx/dy and flip it the other way. Then, I replaced the 'cos^2(y)' and 'sine y' in the final answer with x (by replacing cos y with sqr root (1 - sine^2(y))
Sorry if I confused anyone :l

EDIT: Or you could simply use the function within a function rule. Inverse sine f(x) = f'(x)*(1/sqr root (1-x^2) where f'(x) in this case is -1/x^2
 
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michaeljennings

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its right except for the x^2 at the front and the negative sign..im pretty sure
You have to remember that you have to times the whole thing by the derivative of whats in the brackets to begin with. The derivative of 1/x is -1/x^2 thats why its there
 
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You are right except you had to state that x>1 or x<-1... and you can further simplify it to which is what one of the post above have mentioned.
 

Lindurr

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Thanks for the working out seductive! Thats the answer :)
 

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