derivatives?

Question

Hello. I have been trying to solve this problem for so long but it's just too hard. I don't understand it.

Does anyone know how to find the 1st derivative and 2nd derivative of √2x*sinx (the square root of 2x times sinx. using the 1st and 2nd derivative tests.

I know that I have to use the product rule with the chain rule but it is not working and my teacher said my answer is wrong.

Any answers are appreciated, I just want to understand how to do it.

Thank you!

The equation is: √(2x)*sinx

Answer 1

Chain rule should work in this case.
(√2x*sinx )' = (√2x)' *sinx + √2x*(sinx)' = (1/√2x)*sinx + √2x*cosx
(√2x*sinx )'' = (1/√2x)' *sinx + (1/√2x)*(sinx)' + (√2x)'*cosx + (√2x)*(cosx)' = (2x)^(-3/2)*sinx + (1/√2x)*(cos x) + (1/√2x)* cosx-(√2x)*sinx

Answer 2

f(x) = ((2x)^(1/2))sinx
f'(x) = ((2x)^(1/2))cosx + ((2x)^-(1/2))sinx
f''(x) = - ((2x)^(1/2))sinx + ((2x)^-(1/2))cosx + ((2x)^-(1/2))cosx - ((2x)^-(3/2))sinx
f''(x) = 2((2x)^-(1/2))cosx - (1 + 2x)((2x)^(1/2))sinx