Derive ln(tanx)

Howdo I do this? I tried the chain ruled but did not get the right answer

Question

Derive ln(tanx)

Howdo I do this? I tried the chain ruled but did not get the right answer

amistre64 · Answer

can you show what you did?

amistre64 · Answer

recall that ln(u) derives to u'/u

jdoe0001 · Answer

$\bf \cfrac{d}{dx}[{\color{blue}{ ln[{\color{brown}{ tan(x)}}]}}]\implies {\color{blue}{ \cfrac{dy}{dx}}}\cdot {\color{brown}{ \cfrac{dy}{dx}}}$

anonymous · Answer

this is what I did.

u=tanx

ln(u)*u'

so I got    sec^2x/tan(x)

amistre64 · Answer

sec^2/tan is good ..... it can be reworked of course, but its fine

jdoe0001 · Answer

yeap.... maybe what you're looking at is some simplified version of that

amistre64 · Answer

$$\frac1{c^2}\frac{c}{s}=\frac{1}{cs}$$

amistre64 · Answer

sec cos  maybe?

amistre64 · Answer

lol,  csc not cos   inmy mind it was right but my fingers hate me

jdoe0001 · Answer

yeap.... could be rewritten as csc(x)sec(x)

aum · Answer

What is the "right answer" that you are referring to?

amistre64 · Answer

hope its not x^2 :)

anonymous · Answer

in my book it says the answer is 2csc2x

amistre64 · Answer

well

sin(x) cos(x) = 2sin(2x)

but then thats csc(2x)/2

aum · Answer

$$
\frac{\sec^2(x)}{\tan(x)} = \frac{1}{\cos^2(x)} * \frac{\cos(x)}{\sin(x)} = 
\frac{1}{\sin(x)\cos(x)} = \frac{2}{2\sin(x)\cos(x)} = \frac{2}{\sin(2x)} = 2\csc(2x)
$$

amistre64 · Answer

so what your saying is trig is a horrible subject and should simply be eliminated from all cirrculum .... i agree :)

anonymous · Answer

help me I am confused! how do I get that

amistre64 · Answer

you have to remember your trig indentities