My overarching question is about differentiating when you have these inverse trig functions, but listed below is the specific question I am trying to solve. If you help me with the problem, it'll help me (and others) apply it to similar questions.
Problem: y = arcsin(x) - sqrt(1 - x^2) Find dy/dx
Answer Choices: 1/2sqrt(1-x^2) or 2/sqrt(1-x^2) or (1+x)/sqrt(1-x^2) or (x^2)/sqrt(1-x^2) or 1/sqrt(1+x)
The arcsin(x) is primarily what is getting me stuck. To try to solve the problem I moved the root to the other side by adding it to both sides.
- y + sqrt(1 - x^2) = arcsin(x)
Then I converted the equation into a sin equation...I don't feel like this is correct
- sin(y + sqrt(1 - x^2)) = x
From here, if I take dy/dx of both sides, it seems utterly confusing and on the wrong track. (I believe I applied chain rule correctly, but I could be wrong)
- cos(y + sqrt(1 - x^2)) * [dy/dx + 1/2sqrt(1-x^2) * -2x] = 1
I also examined the square root in the problem carefully because I noticed it had a striking resemblance to another problem I saw earlier in a book:
Differentiate y = arcsin(x)
1. sin(y) = x
2. cos(y) dy/dx = 1
3. dy/dx = 1/cos(y) = 1/sqrt(1- (sin(y))^2) = 1/sqrt(1-x^2) because of the trig identity (sin(x))^2 + (cos(x))^2 = 1 & because subtracting (sin(y))^2 is the same as subtracting x^2 because of Step One conversion
So because I saw the sqrt(1-x^2) in the tougher problem I'm doing right now, I tried to find a way to utilize the technique from the earlier one, but I couldn't. So perhaps that could be the key to solving it.
Thanks in advance for your help, I appreciate it.
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