Question

2x^2tan(x)/sec(x)

find f'(x)

and f'(3)

Answer 1

Are the tangent and secant both in the exponent? :O

Answer 2

\[\frac{ 2x^2tanx }{ secx }\]
Errr I suppose it's like this? :D that would make more sense.

Answer 3

yeah thats it, for some reason my open study isnt working right, so i hope it works because I really need the help

Answer 4

If we try to differentiate it from here, it'll be a bit of a mess, because we have the product rule within the quotient rule...
Rewrite tangent and secant in terms of sines and cosines, and you should see a nice cancellation. :D
Remember the identities?

Answer 5

sec=1/cos and tan=sin/cos?

Answer 6

mhm c:

Answer 7

So you're dividing by secant, when you change it, using that identity, you end up dividing by a fraction. Soooo what can we do with that? :D Remember how to flip it? XD

Answer 8

haha yup multiply it by cos/1

Answer 9

Anything cancel out? :3

Answer 10

the two cos?

Answer 11

yay

Answer 12

\[\frac{ 2x^2tanx }{ secx }=2x^2sinx\]
Wow so we were able to simplify it quite nicely :O
So from here you want to give it a try or still confused? :)
Product rule stuffs.

Answer 13

im gonna give her a go.

Answer 14

if we use the product rule (4x^1) and the derivative of sinx is cosx

Answer 15

Mhm sounds right, come up with an answer? :D

Answer 16

so would f'(x)= (4x)(cos(x))

Answer 17

Woops D: You just differentiated them both, uh ohhhhh.
We have the product of 2 functions of x. (2 things with x).

So we have to apply the product rule.
\[(uv)'=u'v+uv'\]
\[[2x^2tanx]'=(2x^2)'tanx+2x^2(tanx)'\]
Understand what you need to do?
The primes signify a place where you still need to take a derivative.

Answer 18

(4x)(tan(x))+2x^2(sec(x))^2 is that what you mean?

sec(x)^2 = derivative of tan(x)

Answer 19

Yes good job! c:

Answer 20

?? is that the answer

Answer 21

Yes.

Answer 22

That answer didnt work in my online assignment. I put it in www.wolframalpha.com and it says f ' (x) = 2 x (x cos(x)+2 sin(x))