Question

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

Answer 1

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.

Answer 2

`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)`