How to find the derivative of trig function with a radical

such as:

y = √x sin x

The answer should be:

dy/dx = √x cosx + [(√x sinx) / 2x]

Why/how?

Thanks so much!

Use the 'Product' rule.

dy/dx = u(dv) + v(du)

Since x^(1/2)*Sin x is a product
differentiating 'x^(1/2) = (1/2)x^-(1/2) = 1/(2x^(1/2))
differentiating Sin x = Cos x
So
dy/dx = x^(1/2)Cos x + sin x(1/2x^(1/2))
dy/dx = x^(1/2)Cosx + Sin x / 2x^(1/2)

Since x^(1/2) as the denominator may be 'irrational', its conjugate is applied.

Hence
dy/dx = x^(1/2)Cosx + x^(1/2)Sin x / 2x^(1/2)(x^(1/2))
dy/dx = x^(1/2)Cosx + x^(1/2)Sin x / 2x As required!!!!!

You have to do product rule with chain rule. So you would get

y'(x) = the derivative of x^(1/2) * sinx + the deriviative of sinx * sqrt (x)

The derivative of the sqrtx is (1/2) x ^(-1/2) and the derivative of sinx = cosx, so I believe that the answer should be

y'(x) = sinx/ 2*sqrt(x) +sqrt(x) * cosx

confronted with a radical the simplest way to proceed is to think of √x as

x^.5

and solve by the usual power rule d/dx x^u is u*x ^ (u-1) ( if u <> 0)

so,

dy/dx(√x sin x) is

dy/dx(x^.5 sin x)

and by the product rule becomes

sinx *(x^-.5)/2 + cos x * x^(.5)