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In the problem below (step 3), I can't figure out why I'm missing a $(2)$ in $\frac{e^{2x}(2)(2x)-e^{2x}*2}{2x^2}$. Instead, I came up with $\frac{e^{2x}(2x)-e^{2x}*2}{2x^2}$. What did I do wrong here? Thanks in advance.

Derivative

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    $\begingroup$ Hint: What's the derivative of $e^{2x}$? $\endgroup$ – PM 2Ring Jan 4 '17 at 7:13
  • $\begingroup$ I got $2e^{2x}$ $\endgroup$ – buzzard Jan 4 '17 at 7:14
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    $\begingroup$ So the first term of the numerator has to be $\frac{d(e^{2x})}{dx}(2x) = (2e^{2x})(2x)$ $\endgroup$ – PM 2Ring Jan 4 '17 at 7:19
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$$\frac{d}{dx}\frac{e^{2x}}{2x}$$ $$=\frac{2x\frac{d}{dx}e^{2x}-e^{2x}\frac{d}{dx}2x}{(2x)^2}$$.

Now,you have made a mistake in the derivative of $e^{2x}$ it seems.

Here's how to do it.

Let,$e^{2x}=y$ and $2x=z$.Now,$\frac{dy}{dx}=\frac{dy}{dz}\cdot\frac{dz}{dx}=\frac{d}{dz}e^z\cdot\frac{d}{dx}(2x)=e^z\cdot2=2e^{2x}$ .

So,the derivative evaluates to,

$$\frac{2x\cdot e^{2x}\cdot2-e^{2x}\cdot2}{4x^2}$$.

Hope this helped!!

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