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?
$\endgroup$ 24 Answers
$\begingroup$Is that $e^{5x}\cdot\tan2x$? In that case, we want to use the product rule. Recall that the product rule says that $(fg)'(x)=f'(x)g(x)+g'(x)f(x).$
$\endgroup$ 5 $\begingroup$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
$$e^{5x}(5\tan(2x)+2\sec^2(2x))$$
$\endgroup$ $\begingroup$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) $$\begin{align} f'(x)g(x) + f(x)g'(x) &= 5e^{5x}\tan(2x) + e^{5x}\Box \end{align} $$
$\endgroup$ 2 $\begingroup$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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