Derivative product rule $\frac{d}{dx}\sqrt{xe^{-3x}}$

Question

I have this question in one of my textbooks and no matter what approach I take to solving it the answer is always wrong

$\frac{d}{dx}\sqrt{xe^{-3x}}$

It would be greatly appreciated if someone could explain it to me. ps, I'm new to math stack exchange, so please tell me if am asking the right kind of questions. Thanks :)

The textbook says the answer is:

$\frac{e^{-3x}}{2{\sqrt{x}}}-3{\sqrt{xe^{-3x}}}$

Answer 1

Remember that by chain rule

$$\frac{d}{dx}\sqrt{xe^{-3x}}=\frac{f'(x)}{2\sqrt{f(x)}}=\frac{1}{2 \sqrt{x e^{-3x}}} [e^{-3x}-3xe^{-3x}]=\frac{e^{-3x} (1-3x)}{2 \sqrt{x e^{-3x}}}=\\=\frac{e^{-3x} (1-3x)}{2 \sqrt{x e^{-3x}}}\frac{\sqrt{x e^{-3x}}}{\sqrt{x e^{-3x}}}=\frac{(1-3x)\sqrt{x e^{-3x}}}{2x}$$

We can also put in that form

$$...=\frac{(1-3x)\sqrt{x e^{-3x}}}{2x}=\frac{\sqrt{x e^{-3x}}}{2x}-\frac{3x\sqrt{x e^{-3x}}}{2x}=\frac{\sqrt{e^{-3x}}}{2\sqrt x}-3\frac{\sqrt{x e^{-3x}}}{2}$$

Answer 2

If you want to compute $\frac{dy}{dx}$, then you have that

$$ \frac{1}{2 \sqrt{x e^{-3x}}} [e^{-3x}-3xe^{-3x}]=\frac{e^{-3x} (1-3x)}{2 \sqrt{x e^{-3x}}} $$

Answer 3

Just simplify the function to $\sqrt x \cdot e^{-1.5x}$. Then the derivative is $$\frac{1}{2\sqrt{x}}\cdot e^{-1.5x}+\sqrt{x}e^{-1.5x}(-1.5).$$