Differentiation with the quotient rule I have the question:
Given that $$y=\frac{e^x}{\sqrt{1+2x}}$$
Show that $$\frac{dy}{dx} = \frac{2xe^x}{\sqrt{(1+2x)^3}}$$
I've done the question but I got $2xe^x(\sqrt{(1+2x)^3})$. I feel like this is too similar to the sheet's answer, so am I wrong or is the sheet printed wrong?
Working:
Let $u = e^x$, $v = 1+2x$, $w = \sqrt{v}$.
Using the chain rule, $$\frac{dw}{dx} = \frac{dw}{dv} * \frac{dv}{dx}$$
So $$\frac{dw}{dx} = 2(\frac{1}{2}v^{-\frac{1}{2}}) = \frac{1}{\sqrt{v}} = \frac{1}{\sqrt{1+2x}}$$
And $\frac{du}{dx}$ is clearly $e^x$.
Therefore, using the quotient rule, $$\frac{dy}{dx} = \frac{e^x(\sqrt{1+2x})- e^x(\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\sqrt{1+2x}-\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\frac{2x}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{\frac{2xe^x}{\sqrt{1+2x}}}{(1+2x)^2}$$
Using the rule
$$\frac{x^m}{x^n} = x^{m-n}$$
We get
$$\frac{2xe^x}{(1+2x)^{-\frac{3}{2}}}$$
$$=2xe^x(\sqrt{1+2x})^3 $$
Have I done this correctly? 

I suspect 
$$\frac{1}{\left(\sqrt{1+2x}\right)^2} \not = \frac{1}{\left({1+2x}\right)^2}$$
and $$\frac{\frac{1}{\sqrt{1+2x}}}{(1+2x)^2} \not =\frac{1}{(1+2x)^{-\frac{3}{2}}}$$
but $$\frac{\frac{1}{\sqrt{1+2x}}}{\left(\sqrt{1+2x}\right)^2}  = \frac{1}{\left(\sqrt{1+2x}\right)^3}$$ 
