$\displaystyle f(x)=tg x \cdot e^{3x}$ $\displaystyle f'(x)=(tg x \cdot e^{3x})'=...$ What rule I can use to solve this? I need examples like this tasks or something
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Are t and g considered constant?
It's tg(x)
Is that the tangent function, then?
Originally Posted by Ackbeet Is that the tangent function, then? Yes.
Ah. Standard notation is $\displaystyle (\tan(x)e^{3x})'.$ What rule do you think would be appropriate here?
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...?
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.
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 $\displaystyle (u\cdot v)'=u'v+uv'$
Right.
It's correct: $\displaystyle (tan(x))'=\frac{1}{cos^2x}$ $\displaystyle (e^{3x})'=e^{3x}*3$
Correct. So you get what?
$\displaystyle =\frac{1}{cos^2x}\cdot e^{3x}+tgx\cdot e^{3x}\cdot 3$ What do next I don't know?
How about factoring out the $\displaystyle e^{3x}?$
Originally Posted by Ackbeet How about factoring out the $\displaystyle e^{3x}?$ $\displaystyle 3e^{3x}$? How first e^{3x} I don't know