Find dydx for y=sin−1(1x√) .

Answer 1

To find dy/dx for y = sin^(-1)(x^(1/2)), we can use the chain rule.

Let u = x^(1/2).
Then y = sin^(-1)(u).

Taking the derivative of both sides with respect to x:

dy/dx = d/dx (sin^(-1)(u))

Applying the chain rule, we have:

dy/dx = (dy/du) * (du/dx)

To find (dy/du), consider y = sin^(-1)(u) as y = sin^(-1)(x^(1/2)).

Using implicit differentiation, we have:

1 = (dy/dx)*(d/dx)(x^(1/2))

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

Therefore, (dy/dx) = 2*(x^(-1/2)).

Now, let's find (du/dx).

u = x^(1/2)
Taking the derivative of u with respect to x:
du/dx = (1/2)*(x^(-1/2))

Now, we can substitute (dy/du) and (du/dx) back into the chain rule equation:

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

Therefore, dy/dx = 1/x.