# Are my derivatives worked correctly?

I have 3 derivatives that I need to find:

1: $y = 3sin(x) - ln(x) + 2x^\frac{1}{2}$

$3cos(x) - \frac{1}{x} + (\frac{1}{2}) (2x^\frac{-1}{2})$

2: $y = 3e^{2x} * ln(2x + 1)$

$3e^{2x} (\frac{2}{2x+1}) + 6e^{2x} (ln(2x+1))$

3: $y = \frac{sin(2x)}{3x^2 + 5}$

$\frac{(3x^2 + 5)(2sin(2x)) - sin(2x)(6x)}{(3x^2 + 5)^2}$

I'm relatively positive I did these correctly, but please let me know if I haven't.

EDIT

3: $\frac{(3x^2 + 5)(2cos(2x)) - sin(2x)(6x)}{(3x^2 + 5)^2}$

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 In 3: $(\sin{t})'=\cos{t}$ – M. Strochyk Dec 12 '12 at 20:02 Seems I overlooked that, thanks! Glad someone felt the need to downvote the question when I'm genuinely asking just to be checked so I know I'm ready for the exam. Aside from that the community has been quite nice about my questions. – StrugglingWithMath Dec 12 '12 at 20:06 When differentiating $\sin(2x)$ in 3., you use the chain rule and differentiate $\sin$ first. Also, you would (might) probably lose a point for not making obvious and easy simplifications in your first two answers. – David Mitra Dec 12 '12 at 20:07 @DavidMitra The professor has been quite nice about the simplifications. While I could simplify it, he just wants to make sure we know the required steps :) – StrugglingWithMath Dec 12 '12 at 20:09

In $(3)$: $\cos(2x)$ appears nowhere in your solution. That should tell you something about your solution to $(3)$.
You got it right in $(1)$, so I'm assuming you know that the derivative of $\sin(2x) = 2 \cos(2x)$.
Spend some more time on $(3)$, and see what you arrive at.
I'm assuming you have written intermediate steps, and simply posted your solutions. Be sure that if this is homework, that you hand in a more detailed, step-by-step solution. You can simplify a little bit, too, e.g., in $(1)$.