$\displaystyle =\frac{-2}{\sqrt{2-4x^2}}$

Answers says its:$\displaystyle \frac{1}{\sqrt{x-x^2}}$

2)$\displaystyle y=sin^{-1}(\sqrt{x})$

$\displaystyle \frac{dy}{dx}=$$\displaystyle \frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})$

$\displaystyle =\frac{1}{2\sqrt{x}(\sqrt{1-x^2})}$

Answers says its:$\displaystyle \frac{1}{2\sqrt{x-x^2}}$

3)Find the second derivative of $\displaystyle xe^{2x}.$

The first derivative is $\displaystyle e^{2x}(1+2x)$

The second derivative i got:

$\displaystyle =2e^{2x}+4xe^{2x}$

$\displaystyle =2e^{2x}(1+2x)$

Answers says its $\displaystyle 4(x+1)e^{2x}$

4)If $\displaystyle y=2cos3x+5sin3x$, show that$\displaystyle \frac{d^2y}{dx^2}=-9y$

First derivative i got: $\displaystyle -6sin3x+15cos3x$

Now second derivative i got: $\displaystyle -18cos3x-45sin3x$

What did i do wrong? or what is the next step?

P.S