finding f prime

Question

finding f'(1/4)

let f(x) = (arcsin(√x))² find f'(1/4)

Answer 1

Expert Al L's answer is very correct.

f(x)=(sin^-1(x^(1/2))²

f'(x)=2(sin^-1(x^(1/2))(1/(1-x)^(1/2))(1/2)x^(-1/2),,,,using chain rule,,,,d(sin^-1z(x))/dx=(dz/dx)/(1-z²)^(1/2)

f'(x)=(sin^-1(x^(1/2)))(1/(x-x²)^(1/2))

note √1/4)=1/2 and sin^-1(1/2)=pi/6

Answer 2

f(x)=(arcsin(√x))²

f'(x)=2(arcsin(√x))•d/dx[arcsin(√x)] (by the chain rule)

→f'(x)=2(arcsin(√x))•(1/√(1-(√x)²))•d/dx[√x] (by the chain rule)

→f'(x)=2(arcsin(√x))•(1/√(1-x))•(1/(2√x))

→f'(x)=(arcsin(√x))/(√(1-x)•√x) (if this step is difficult to see I encourage you to write out the equation on paper and see how I have simplified the equation to this step.

@ x = 1/4

f'(1/4)=arcsin(√(1/4))/(√(1-(1/4))•√(1/4))

f'(1/4)=arcsin(1/2)/(√(3/4)*(1/2))

f'(1/4)=arcsin(1/2)/(√3/2*1/2)

f'(1/4)=(π/6)/(√3/4)

f'(1/4)=4π/(6√3)

Please feel free to ask any questions