Find dydx for y=sin^−1(1√x) .

Answer 1

To find dy/dx for the equation y = sin^(-1)(1/√x), we can use the chain rule.

Let u = 1/√x, then y = sin^(-1)(u).

The derivative of sin^(-1)(u) with respect to u is 1/√(1-u^2).

Now, we need to find the derivative of u with respect to x.

Taking the derivative of u = 1/√x with respect to x, we get du/dx = -1/2x^(3/2).

Multiplying these two derivatives together using the chain rule, we have:

dy/dx = (1/√(1-u^2)) * (-1/2x^(3/2))

Since u = 1/√x, we can substitute it back into the equation:

dy/dx = (1/√(1-(1/√x)^2)) * (-1/2x^(3/2))

Simplifying further:

dy/dx = (1/√(1-1/x)) * (-1/2x^(3/2))

= (-1/2x^(3/2)) * (1/√(x-1)/x)

= -(1/2x^(3/2)) * (x/√(x-1))

= -(x/2x^(3/2)) * (1/√(x-1))

= -x/(2√x * √(x-1))

Therefore, the derivative dy/dx for y = sin^(-1)(1/√x) is -x/(2√x * √(x-1)).