Question

What is the derivative of `ArcSin[x^(1/2)]`?

Answer 1

`:. dy/dx = 1/(2sqrt(x)sqrt(1-x))`

Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you use the chain rule.

Let `y=arcsin1(x^(1/2)) <=> siny=sqrt(x)`

Differentiate Implicitly:

`cosydy/dx = 1/2x^(-1/2)`
`:. cosydy/dx = 1/(2sqrt(x))` ..... [1]

Using the `sin"/"cos` identity;

`sin^2y+cos^2y -=1`
`:. cos^2y=1-sin^2y`
`:. cos^2y=1-(sqrt(x))^2`
`:. cos^2y=1-x`
`:. cosy=sqrt(1-x)`

Substituting into [1]
`:. sqrt(1-x)dy/dx = 1/(2sqrt(x))`
`:. dy/dx = 1/(2sqrt(x)sqrt(1-x))`