Differentiate $\frac{e^{-2x}}{\sqrt x}$ 
Differentiate, with respect to $x$, $\frac{e^{-2x}}{\sqrt x}$.

I  a having difficulties with differentiation. The answer is
$$-\frac{e^{-2x}(4x+1)}{2x\sqrt{x}}$$
But I am looking at my working out and can't seem to solve the question. This is what I did:
$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}\times \frac{d}{dx}(e^{-2x})-e^{-2x}\times \frac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2} = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}\frac12(\sqrt{x})^{-1/2}}{x}$$
$$ = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}}{2x\sqrt{x}}$$
image of my work
If someone could please help me solve this question or give advise please help me! Thank You!
 A: We have $f(x)=\dfrac{e^{-2x}}{\sqrt x}$.
Use Quotient rule as you suggested in your answer too:
$$f'(x)=\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}$$
Now multiply the fraction by $\frac{2\sqrt x}{2\sqrt x}$ to get rid of $\frac{-1}{2\sqrt x}$ in numerator:
$$\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}\times\frac{2\sqrt x}{2\sqrt x}=\frac{-4e^{-2x}\times x-e^{-2x} }{2x\sqrt x}=-\frac{e^{-2x}(4x+1)}{2x\sqrt x}$$
A: Note that $\frac{1}{\sqrt x} = x^{-1/2}$. Now use the product rule:
$$\frac{d}{dx} \left(e^{-2x} \cdot x^{-1/2}  \right)= e^{-2x} \cdot -\frac{1}{2}x^{-3/2} + (e^{-2x} \cdot -2) \cdot x^{-1/2}$$
which is already correct, but can be written in a nicer form:
$$=e^{-2x} \left(-\frac{1}{2}x^{-3/2} - 2x^{-1/2} \right)=e^{-2x} x^{-3/2} \left(-\frac{1}{2} - 2x \right)= -\frac{1/2 + 2x}{e^{2x} x^{3/2}} = -\frac{1+4x}{2e^{2x} x^{3/2}}$$
A: You need to use $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{d}{dx}u-u\frac{d}{dx}v}{v^2}.$$
When this formula is compared to your question, you have $u = e^{-2x}$ and $v=\sqrt{x}.$ Just substitute and simplify.
That is $$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}\frac{de^{-2x}}{dx}-e^{-2x}\frac{d\sqrt{x}}{dx}}{(\sqrt{x})^2}.$$
$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}e^{-2x}(-2)-e^{-2x}\frac{1}{2\sqrt{x}}}{(\sqrt{x})^2}.$$
After simplification, one should get
$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{-4xe^{-2x}-e^{-2x}}{2x\sqrt{x}} = -\frac{e^{-2x}(4x+1)}{2x\sqrt{x}} $$
