# Calculate f'(x)

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• October 21st 2010, 10:26 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, 10:27 AM
Ackbeet
Are t and g considered constant?
• October 21st 2010, 10:28 AM
Lil
It's tg(x)
• October 21st 2010, 10:31 AM
Ackbeet
Is that the tangent function, then?
• October 21st 2010, 10:33 AM
Lil
Quote:

Originally Posted by Ackbeet
Is that the tangent function, then?

Yes.
• October 21st 2010, 10:34 AM
Ackbeet
Ah. Standard notation is

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

What rule do you think would be appropriate here?
• October 21st 2010, 10:36 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, 10:37 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, 10:45 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, 10:46 AM
Ackbeet
Right.
• October 21st 2010, 10:55 AM
Lil
It's correct:
$(tan(x))'=\frac{1}{cos^2x}$
$(e^{3x})'=e^{3x}*3$
• October 21st 2010, 10:56 AM
Ackbeet
Correct. So you get what?
• October 21st 2010, 11:00 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, 11:01 AM
Ackbeet
How about factoring out the $e^{3x}?$
• October 21st 2010, 11:05 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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