Finding 1st and 2nd 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!

This is the equation: √(2x)*sinx

Answer 1

Like MD said, it's not clear if you mean (√2x) times sin x, or (√2)x times sin x, or √(2x sin x).

d/dx (√2x) sin x
= (d/dx √2x) sin x + (√2x) (d/dx sin x) (product rule)

d/dx √(2x sin x)
= d/dx (2x sin x)^(1/2) (^ means exponentiation)
= (1/2) (2x sin x)^(-1/2) d/dx(2x sin x) (chain rule)
= (1/2) (2x sin x)^(-1/2) ((d/dx 2x) sin x + 2x (d/dx sin x)) (product rule)

You can do the last steps yourself.

Answer 2

I don't know what your answer was, so I can't comment on whether it was right or wrong.

y = √(2x) sin x
dy/dx = d/dx (√(2x)) . sin x + √(2x) . d/dx (sin x) - product rule
= (1/2) (2x)^(-1/2) (2) . sin x + √(2x) cos x - chain rule
= sin x / √(2x) + cos x √(2x)

d^2y/dx^2 = d/dx [(sin x) (2x)^(-1/2) + (cos x) (2x)^(1/2)]
- product and chain rule again:
= (cos x) (2x)^(-1/2) + (sin x) (-1/2) (2x)^(-3/2) (2) - (sin x) (2x)^(1/2) + (cos x) (1/2) (2x)^(-1/2) (2)
= cos x / √(2x) - sin x / (2x)^(3/2) - (sin x) √(2x) + cos x / √(2x)
= [4x cos x - sin x - 2x sin x] / (2x)^(3/2)

Answer 3

Please clarify if it is
√2x*sinx
OR
√(2x)*sinx
OR
√(2x*sinx)
OR
√2(x) * (sinx)