Question

Find the second derivative of xtanx. So for this one, the first derivative I got is xtan(x)^2+1. So then for the second derivative, I used product rule using x as one product and tan(x)^2+1 as the second product. So the answer to the question is [2sec(x)^2](xtanx+1). And I get really close to that, but for some reason I get [2sec(x)^2][xtan(x)^2+1] instead of [2sec(x)^2](xtanx+1).

Answer 1

Hmm ok let's see what's going on.

Answer 2

\[\Large (x \tan x)'\quad=\quad \color{royalblue}{(x)'}\tan x+x\color{royalblue}{(\tan x)'}\]
\[\Large =\quad \tan x+x \sec^2x\]

Answer 3

Hmmmm that looks a bit different than your first derivative :o
Did you product rule for the first deriv?

Answer 4

No i didn't! okay that's where I went wrong. so let me use that and try for the second derivative and see if i get the answer!

Answer 5

hah that fox so silly c:

Answer 6

ok so i used the new derivative and so this is my process\[\sec(x)^{2}+[(x')\sec(x)^{2}+x(\sec'(x)^{2})]\] i just took the derivative of tan and then used product on the second half of the equation. so I think you use chain on the \[\sec(x)^{2}\]and that should be 2sec(x)*tan(x)*sec(x) right? and but I've tried simplifying it and moving it around, but i still don't get the answer. did the silly fox make a mistake again lol?

Answer 7

Yah you've got the right parts now, just simplification throwing you off?\[\Large \sec^2x+\sec^2x+2x \sec x (\sec x \tan x)\]

Answer 8

ok so I'm simplifying it and so I have the 2sec(x)^2 but im not sure how to simplify \[2xsec(x)[\sec(x)\tan(x)]\] into xtan(x)+1.
sorry im so bad at this, but thanks a bunch for helping thus far!!

Answer 9

Combined the first two terms? Ok great.\[\Large 2\sec^2x+\color{orangered}{2x \;\sec x (\sec x \tan x)}\]For the other term let's multiply the secants together,\[\Large 2\sec^2x+\color{orangered}{2x \;\sec^2 x (\tan x)}\]

Answer 10

and so then the sec(x)^2 can become tan(x)^2+1 right?

Answer 11

Nono we don't want to do that.
See how each term contains a sec^2x?

*cough* factor *cough cough* :o

Answer 12

ooh okay so should it be \[2\sec ^{2}x[xtan(x)]\] or just \[\sec ^{2}x[2+2x(\tan(x)]\]

Answer 13

The second form looks correct.
Looks like you're having a little bit of trouble factoring the 2 out correctly.

So look at the second form that you posted.
If we factor a 2 out of 2, what does leave us in it's place within the brackets?
(Hint: it's not zero)

Answer 14

OOOOOOOOOOOOOHHHHHHHHHHHHHHHHHHHHH *realization hits like a brick wall* You get the answer!!!!

Answer 15

lol XD