How to solve this implicit differentiation problem concerning arcsin?

Question

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.

1. y + sqrt(1 - x^2) = arcsin(x)

Then I converted the equation into a sin equation...I don't feel like this is correct

2. 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)

3. 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.

Answer 1

1. y + sqrt(1 - x^2) = arcsin(x)
2. sin(y + sqrt(1 - x^2)) = x
3. cos(y + sqrt(1 - x^2)) * [dy/dx + 1/2sqrt(1-x^2) * -2x] = 1
4. cos(arcsin(x)) * [dy/dx - x/sqrt(1-x^2)] = 1
5. sqrt(1-x^2) * [dy/dx - x/sqrt(1-x^2)] = 1
6. dy/dx = 1/sqrt(1-x^2) + x/sqrt(1-x^2)
7. dy/dx = (1+x)/sqrt(1-x^2)