Question

f(x)=[2x^2tan(x)]/sec(x)
find the f'(x) and f'(3)

Answer 1

\[f(x)=\frac{ 2x^2tanx }{ secx }\]
Does that look right? c: I couldn't tell if the tangent was suppose to be in the exponent or not heh.

Answer 2

what you wrote is what the problem is

Answer 3

use the quotient rule

Answer 4

i don't know how to differentiate the sec(x)

Answer 5

maybe rewrite each trig term in terms of sines and cosines :) that might make it a lil simpler

Answer 6

(f/g)' = g f' - f g'
--------
g^2

Answer 7

derivative of sec x = sec x tan x

Answer 8

i got
\[4xsec^2*secx-secx*\tan*2x^2tanx / \sec2x\]
but it's not right

Answer 9

So unfortunately for this problem, you have the product rule WITHIN the quotient rule :C
So this is going to be a bit of work D':

Answer 10

@ brooke_army check out your question on Calculus 1

Answer 11

the answer is 2x( x cos x + 2 sinx)

Answer 12

the steps are in Calculus 1

Answer 13

r u there?