How do you find the derivative of F(x)=arcsin sqrt sinx?

2 Answers
Apr 7, 2017

Use the chain rule to find the derivative, because your function F(x) can be expressed as G(h(j(x))) where G(x) = arcsin(x), h(x) = x^(1/2) and j(x) = sin(x)

Explanation:

The chain rule states that the derivative of a composite function is equal to the derivative of the individual composite functions with respect to their inner functions multiplied by the inner functions derivative with respect to x.

So F'(x) = G'(h(j(x)) * h'(j(x)) * j'(x)

Since G'(x) = (1/(sqrt(1-x^2))), G'(h(j(x))) = 1/(sqrt(1 - (sqrt(sin(x)))^2)
And h'(x) = 1/2sqrt(x) , h'(j(x)) = 1/2(sqrt(sin(x)))
And j'(x) = cos(x)

Thus the derivative becomes (1/(sqrt(1 - (sqrt(sin(x)))^2))) * (1/(2(sqrt(sin(x))))) * (cos(x))

Simplifying, we get (cos(x))/(2*(sqrt(1-sin(x))) * (sqrt(sin(x))))

This answer can be further simplified using trigonometric identities.

Apr 7, 2017

F'(x) = cosx/(2 sqrtsinx sqrt(1-sinx)

Explanation:

F(x)=arcsin sqrt sinx

Take sines of both sides:

sin F(x)= sqrt sinx

Implicit differentiation wrt x#:

cos F(x) cdot F'(x) = 1/2 1/sqrtsinx cosx

Re-arrange:

F'(x) =cosx/(2 sqrtsinx ) cdot 1/ (cos F(x))

= cosx/(2 sqrtsinx) cdot 1/ (cos (arcsin sqrt sinx))

Now draw a right-angled triangle and you will see that:

cos ( arcsin theta) = sqrt(1-theta^2), or arcsin theta = arccos( sqrt(1-theta^2))

implies F'(x) = cosx/(2 sqrtsinx sqrt(1-sinx)