Calculate f'(x)

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October 21st 2010, 09:26 AM
Lil

Calculate f'(x)

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

What rule I can use to solve this? I need examples like this tasks or something
October 21st 2010, 09:27 AM
Ackbeet

Are t and g considered constant?
October 21st 2010, 09:28 AM
Lil

It's tg(x)
October 21st 2010, 09:31 AM
Ackbeet

Is that the tangent function, then?
October 21st 2010, 09:33 AM
Lil

Quote:

Originally Posted by Ackbeet

Is that the tangent function, then?

Yes.
October 21st 2010, 09:34 AM
Ackbeet

Ah. Standard notation is

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

What rule do you think would be appropriate here?
October 21st 2010, 09:36 AM
Also sprach Zarathustra

Yes, mr. Ackbeet it is tan(x)....

Appalling the chain rule...

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

Could you proceed from here...?
October 21st 2010, 09:37 AM
Ackbeet

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.
October 21st 2010, 09:45 AM
Lil

Quote:

Originally Posted by Ackbeet

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.

ok, it's like $(u\cdot v)'=u'v+uv'$
October 21st 2010, 09:46 AM
Ackbeet

Right.
October 21st 2010, 09:55 AM
Lil

It's correct:
$(tan(x))'=\frac{1}{cos^2x}$
$(e^{3x})'=e^{3x}*3$
October 21st 2010, 09:56 AM
Ackbeet

Correct. So you get what?
October 21st 2010, 10:00 AM
Lil

$=\frac{1}{cos^2x}\cdot e^{3x}+tgx\cdot e^{3x}\cdot 3$

What do next I don't know?
October 21st 2010, 10:01 AM
Ackbeet

How about factoring out the $e^{3x}?$
October 21st 2010, 10:05 AM
Lil

Quote:

Originally Posted by Ackbeet

How about factoring out the $e^{3x}?$

(Doh) $3e^{3x}$ ?
How first e^{3x} I don't know

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