Question

How do you differentiate `y=sin^-1sqrtx`?

Answer 1

`dy/dx = 1/(2sqrtxsqrt(1 -x))`

Explanation:

`y= sin^-1sqrtx <=> siny=x^(1/2)`

Differentiating (Implicitly) wrt `x`:

`cosydy/dx = 1/2x^(-1/2)`
`:. cosydy/dx = 1/(2sqrtx)`

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

And so:
`:. sqrt(1 -x)dy/dx = 1/(2sqrtx)`
`:. dy/dx = 1/(2sqrtxsqrt(1 -x))`