Question

Find the derivative.
f(x)= tan(3x)(x^2+1)

Answer 1

\[\Large\bf\sf f(x)\quad=\quad (x^2+1) \tan(3x)\]Derivative?
Ok we have the `product` of two x-thingies.
Product indicates we should apply a certain rule, yes? Hmmmmm

Answer 2

So you'd use the (f'(g(x))(g'(x))?

Answer 3

That's called `chain rule`.
Yes we will eventually need to use that.
But that's not what we want to start with.

Our function consists of a product of functions of x.
So we need to start by applying the `product rule`.\[\Large\bf\sf f'(x)\quad=\quad \color{royalblue}{\left[x^2+1\right]'}\tan(3x)+(x^2+1)\color{royalblue}{\left[\tan(3x)\right]'}\]

Answer 4

That's our setup for product rule.
We need to take the derivative of the blue parts.

Answer 5

So the derivative would be 2x for the first one and 3sec^2x?

Answer 6

Oh oh one small boo boo..

Answer 7

*eats zepdrix's head and lets go*

Answer 8

\[\Large\bf\sf f'(x)\quad=\quad \color{orangered}{\left[2x\right]}\tan(3x)+(x^2+1)\color{orangered}{\left[3\sec^2(\color{royalblue}{x})\right]}\]

Answer 9

We need to fix that part in blue.

Answer 10

Ahh my head D:

Answer 11

missing 3x

Answer 12

When you take the derivative of the `outer function` (tangent), the inner function should NOT change. right?