Question

first, second derivative of

Answer 1

also the third derivative

Answer 2

how far did you get ?

Answer 3

first derivative of lncosx is secx right? 1/cosx

Answer 4

for 2 tanx.. the first derivative would be 2sec^2x

Answer 5

d ln x = 1/x dx
in your case x is cos(x)
so dx is d(cos(x))
i.e. use the chain rule

Answer 6

cosxsinx?

Answer 7

d ln cos x = 1/cos x (-sin x) dx

if we take the derivative with respect to x:
\[ \frac{d}{dx} \ln \cos x = \frac{1}{\cos x} -\sin x \frac{dx}{dx} \\
= - \frac{\sin x}{\cos x}
\]

Answer 8

= - tan x

Answer 9

yes -tanx

Answer 10

-tanx is the third derivative?

Answer 11

-tan x is the first derivative of ln cos x

Answer 12

now find the derivative of - tan x to get the 2nd derivative.

Answer 13

-sec^2x

Answer 14

yes.
to find the 3rd derivative of ln cos x
find the derivative of
\[- \sec^2 x \]

Answer 15

2tansec^2x

Answer 16

I forgot -

Answer 17

yes. that was a bit tricky.

Answer 18

How about 2tanx? I got 2sec^2x

Answer 19

sorry u r correct

Answer 20

d/dx 2 tan x = 2 sec^2 x