$\displaystyle f(x)=tg x \cdot e^{3x}$

$\displaystyle f'(x)=(tg x \cdot e^{3x})'=...$

What rule I can use to solve this? I need examples like this tasks or something

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- Oct 21st 2010, 09:26 AMLilCalculate f'(x)
$\displaystyle f(x)=tg x \cdot e^{3x}$

$\displaystyle f'(x)=(tg x \cdot e^{3x})'=...$

What rule I can use to solve this? I need examples like this tasks or something - Oct 21st 2010, 09:27 AMAckbeet
Are t and g considered constant?

- Oct 21st 2010, 09:28 AMLil
It's tg(x)

- Oct 21st 2010, 09:31 AMAckbeet
Is that the tangent function, then?

- Oct 21st 2010, 09:33 AMLil
- Oct 21st 2010, 09:34 AMAckbeet
Ah. Standard notation is

$\displaystyle (\tan(x)e^{3x})'.$

What rule do you think would be appropriate here? - Oct 21st 2010, 09:36 AMAlso sprach Zarathustra
Yes, mr. Ackbeet it is tan(x)....

Appalling the chain rule...

$\displaystyle f'(x)=(tag x \cdot e^{3x})'= (tan(x))'e^{3x}+tan(x)(e^{3x})'$...

Could you proceed from here...? - Oct 21st 2010, 09:37 AMAckbeet
Well, technically, it's the product rule. But yes, that is the rule to use here. I was hoping Lil would see that on her own.

- Oct 21st 2010, 09:45 AMLil
- Oct 21st 2010, 09:46 AMAckbeet
Right.

- Oct 21st 2010, 09:55 AMLil
It's correct:

$\displaystyle (tan(x))'=\frac{1}{cos^2x}$

$\displaystyle (e^{3x})'=e^{3x}*3$ - Oct 21st 2010, 09:56 AMAckbeet
Correct. So you get what?

- Oct 21st 2010, 10:00 AMLil
$\displaystyle =\frac{1}{cos^2x}\cdot e^{3x}+tgx\cdot e^{3x}\cdot 3$

What do next I don't know? - Oct 21st 2010, 10:01 AMAckbeet
How about factoring out the $\displaystyle e^{3x}?$

- Oct 21st 2010, 10:05 AMLil