derivative of trig function

[dlthompson81]

I would greatly appreciate some help here. I am kind of lost.

The problem is to find the second derivative.

f(x) = 3 tan x

I know from the back of the book that the answer is 6sec^2xtanx , but I can't figure out how to get it. Here is what I worked out:

For the first derivative, I used the Product Rule and got:

(3)(sec^2x)+(tanx)(0)
f '(x)= 3sec^2x

Then, I tried to use the Product Rule again to get the second derivative and got:

(3)(sec^2xtan^2x)+(sec^2x)(0)
f ''(x) = 3sec^2xtan^2x

Where did I go wrong?

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ 3tan(x) \ = \ \frac{3sin(x)}{cos(x)}\)

\(\displaystyle f'(x) \ = \ \frac{cos(x)[3cos(x)]-3sin(x)[-sin(x)]}{cos^2(x)} \ = \ \frac{3[cos^2(x)+sin^2(x)]}{cos^2(x)} \ = \ 3sec^2(x), \ Quotient \ Rule\)

\(\displaystyle f"(x) \ = \ 6sec(x)[sec(x)tan(x)] \ = \ 6sec^2(x)tan(x), \ Chain \ Rule.\)