Derivative Check?

Question

Hello. I have been trying to solve this problem and I want to know if it is correct. I had 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.

1st Derivative:
Step 1: √(2x)*sinx
Step 2: find udv + vdu

√(2x)(cosx) + (sinx)(2x)^-1/2 = f’(x) or
√(2x)cosx + (2x)^-1/2sinx

2nd Derivative:
Step 1: √(2x)cosx + (2x)^-1/2sinx
Step 2: find udv + vdu of √(2x)cosx
u= √(2x) v= cosx
du= (2x)-1/2 dv= - sinx
√(2x)(-sinx) + (cosx)(2x)^-1/2 or
-√(2x)sinx + (2x)-^1/2 cosx

Step 3: find udv + vdu of (2x)-1/2sinx

-½(2x)(-1/2)-1 * 2 chain rule


(2x)-1/2(cosx) + (sinx)(-(2x)-3/2) or

(2x)-1/2cosx + (-2x)-3/2sinx

Step 4: Put it together:

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

Any comments are appreciated, I just want to know if I did it right.
Thanks!

Answer 1

This line:

(2x)-1/2(cosx) + (sinx)(-(2x)-3/2)

should read:

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

which makes your last part of this step:

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

If you look above at the end of step 1, you have (2x)^-1/2sinx, but you missed the caret (^) when you started step 3 of your second derivative.

In any event, these changes make your final answer off by a bit, since you're now able to combine a few terms:

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

Otherwise, your process is right on!