Question

Find dy/dx of y=x^2e^xtanx? Please help

Answer 1

\( y = x^2 e^x \tan x \)
Here, we have a product of functions. To find the derivative, we need to apply the product rule.

\( \displaystyle \frac{d}{dx} \left( f(x) g(x) \right) = \frac{df(x)}{dx} g(x) + f(x) \frac{dg(x)}{dx} \)
Of course, we have three factors here. We could either make up a new rule for three factors or we could ues a little trick immediately. I like the latter, although making up the rule will help you see how the product rule generalises.
Let's call \(f(x) = x^2 e^x \) and \(g(x) = \tan x\). We have a product of two functions now, we can apply the first product rule!
\(\displaystyle \frac{d}{dx} \left( \color{#aa0000}{f(x)} \color{#999900}{g(x)} \right) = \frac{d}{dx} \left( \color{#aa0000}{x^2 e^x} \color{#999900}{\tan x} \right) \)
\(\displaystyle \frac{d}{dx} \left( \color{#aa0000}{x^2 e^x} \color{#999900}{\tan x} \right) = \frac{d}{dx} \left( \color{#aa0000}{x^2 e^x} \right) \color{#999900}{\tan x} + \color{#aa0000}{x^2 e^x} \frac{d}{dx} \left( \color{#999900}{\tan x} \right) \)

So, now we have another derivative of a group of factors, but it is only two factors this time! We just need to apply it one more time to break it apart completely. :)

Answer 2

\(\displaystyle \frac{d}{dx} \left( \color{#aa0000}{x^2 e^x} \color{#999900}{\tan x} \right) = \frac{d}{dx} \left( \color{#aa0000}{x^2 e^x} \right) \color{#999900}{\tan x} + \color{#aa0000}{x^2 e^x} \frac{d}{dx} \left( \color{#999900}{\tan x} \right) \)

* \( \displaystyle \frac{d}{dx} \left( \color{#aa0000}{x^2} \color{#770000}{e^x} \right) = \frac{d}{dx} \left( \color{#aa0000}{x^2} \right) \color{#770000}{e^x} + \color{#aa0000}{x^2} \frac{d}{dx} \left( \color{#770000}{e^x} \right) \)

\(\displaystyle \frac{d}{dx} \left( \color{#aa0000}{x^2 e^x} \color{#999900}{\tan x} \right) = \left( \frac{d}{dx} \left( \color{#aa0000}{x^2} \right) \color{#770000}{e^x} + \color{#aa0000}{x^2} \frac{d}{dx} \left( \color{#770000}{e^x} \right) \right) \color{#999900}{\tan x} + \color{#aa0000}{x^2 e^x} \frac{d}{dx} \left( \color{#999900}{\tan x} \right) \)

\(\displaystyle \frac{d}{dx} \left( \color{#aa0000}{x^2 e^x} \color{#999900}{\tan x} \right) =\frac{d}{dx} \left( \color{#aa0000}{x^2} \right) \color{#770000}{e^x} \color{#999900}{\tan x} + \color{#aa0000}{x^2} \frac{d}{dx} \left( \color{#770000}{e^x} \right) \color{#999900}{\tan x} + \color{#aa0000}{x^2 e^x} \frac{d}{dx} \left( \color{#999900}{\tan x} \right) \)

So, if you can see it now, we ended up taking the derivative of each factor once per term. That is a general product rule being applied. That is, if you need to take a derivative of a bunch of products, it should end up like this: ((fghk)' = f' g h k + f g' h k + f g h' k + f g h k' )

All we need to do now is simplify those individual derivatives, which I leave for you to do on your own if you see this. :)