Determine using calculus dy/dx if y =arcsin x/(1-x^2)?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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.