# Determining the derivative of $e^{5x}\tan(2x)$

Determine the derivative of $e^{5x}\tan(2x)$

I have no idea, what is this question mean- but what I can do is
\begin{align*}y&=e^{5x}\\ \frac{dy}{dx}&=e^{5x}(5)\\ \frac{dy}{dx}&=5e^{5x} \end{align*}

what can I do to the $\tan(2x)$. can you please help me out?

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You seem to know the chain rule. Do you know the product rule? Remember that $\frac{d}{dx}(\tan x) = (\sec x)^2$. – Dylan Moreland Jun 2 '12 at 3:14
does it mean - $e^{5x}tan2x+5tan2$ by using product rule. sorry if wrong. – Sb Sangpi Jun 2 '12 at 3:25

Is that $e^{5x}\cdot\tan2x$? In that case, we want to use the product rule. Recall that the product rule says that $(fg)'(x)=f'(x)g(x)+g'(x)f(x).$

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the problem is $e^{5x}.tan2x$ is not $e^{5x}$ x $tan2x$ – Sb Sangpi Jun 2 '12 at 3:23
@SbSangpi What is it, then? You'd have to explain the notation. – Dylan Moreland Jun 2 '12 at 3:25
I don't know, In the question it is like that exactly. It is possible mistake? – Sb Sangpi Jun 2 '12 at 3:26
@SbSangpi, that is why I was asking, I did not recognize that notation as anything meaningful so was assuming that he had intended it to be what I wrote, but wanted to clarify to be sure. – Nathan Bouscal Jun 2 '12 at 3:27
So, it should be $e^{5x}tan2x$! and the question from paper is mistake! thx – Sb Sangpi Jun 2 '12 at 3:29

Assuming your problem is $e^{5x}\tan(2x)$, the derivative would be (using product rule)

$$e^{5x}\cdot\sec^2(2x)\cdot{2}+\tan(2x)\cdot e^{5x}\cdot{5}$$

Which simplifies to

$$e^{5x}(5\tan(2x)+2\sec^2(2x))$$

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You have a product of two functions. $$e^{5x}\tan(2x) = f(x)g(x)$$ where \begin{align} &f(x) = e^{5x} & &g(x) = \tan(2x). \end{align} The product rule says that the derivative of the product (i.e. what you are asked to find) is $$f'(x)g(x) + f(x)g'(x).$$ Hence all you really need to do is to find the derivatives of $f(x)$ and $g(x)$. You already listed that you know that $f'(x) = 5e^{5x}$. Now you just have to look up what the derivative of $\tan(x)$ is. You probably have this written in your textbook. Otherwise you could look here. So $g'(x) = (\tan(2x))' = \Box$ (you fill in what goes in the box $\Box$).

When you have written down the derivative of $f(x)$ and $g(x)$ you get (using the product rule) \begin{align} f'(x)g(x) + f(x)g'(x) &= 5e^{5x}\tan(2x) + e^{5x}\Box \end{align}

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shouldn't $g(x) = \tan(2x)$? – algebra_fan Jun 2 '12 at 3:49
@algebra_fan: Indeed. I wrote it wrong at first, but it is corrected now. – Thomas Jun 2 '12 at 3:52

Well, as $d/dx \tan x = \sec^2 x$, one has $y = 2x \tan 5x$ and so it holds that $y' = 2 \tan5x + 10x \sec^2 x$

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This does not answer the question asked. If you are answering a similar question to show how to approach the question asked, it would be helpful to mention this. – robjohn Apr 7 '13 at 3:19