# Determine using calculus dy/dx if y =arcsin x/(1-x^2)?

### 2 Answers | Add Yours

We have to find the derivative of y = arc sin x/(1-x^2).

We use the quotient rule here:

y' = [(arc sin x)'*(1 - x^2) - ( arc sin x)*(1 - x^2)']/(1 - x^2)^2

=> [sqrt(1-x^2)*(1 - x^2) + 2x*(arc sin x)]/(1 - x^2)^2

**The required derivative dy/dx = [sqrt(1-x^2)*(1 - x^2) + 2x*(arc sin x)]/(1 - x^2)^2**

Since the given function is a fraction, we'll apply the quotient rule to determine the first derivative.

(u/v)' = (u'*v - u*v')/v^2

We'll put u = arcsinx

du/dx = d(arcsinx)/dx = sqrt(1-x^2)

We'll put v = (1-x^2)

dv/dx = d(1-x^2)/dx

dv/dx = -2x

Now, we'll differentiate the function:

dy/dx = [(1-x^2)sqrt(1-x^2) + 2x(arcsinx)]/(1-x^2)^2

**The first derivative of the given function is: dy/dx = [(1-x^2)sqrt(1-x^2) + 2x(arcsinx)]/(1-x^2)^2.**