Question

I have a question about a derivative of trigonometric functions. Details in the reply.

Answer 1

I must show that
\[\frac{d(\ln \sin x)}{d(\ln x)} = x\tan x\]
I've tried it as follows:
\[d(\ln \sin x) = x\tan x\ d(\ln x)\]
But
\[ d(\ln x) = \frac{1}{x}dx \]
So
\[d(\ln \sin x) = x\tan x\ \frac{1}{x}dx\]
\[d(\ln \sin x) = \tan x\ dx\]
which is certainly not true.
What am I doing wrong?

Answer 2

U substitution

Answer 3

they should cancel out after u derive the u subtitution, and u will have an easy derivative

Answer 4

for the ln(sinx) u will have cosx/sinx as the answer...need more help?

Answer 5

which is = to 1/tanx and lnx= 1/x, so the equation is equal to xtanx

Answer 6

What is wrong with the steps I used above?

Answer 7

d/dx ( ln(sinx)) = 1/tanx

Answer 8

I still don't see it correctly.

Answer 9

It also appears to me that x tan x isn't correct. Maybe it is just an error of the book.

Answer 10

Corrected:
d(ln sin x)/d(ln x) = d(ln sin x)/dx . dx/d(ln x)
= d(ln sin x)/dx . [ d(ln x)/dx ]^-1
= cot x / (1/x)
= x / tan x , etc. etc.

I'm confident your proposed answer of x tan x isn't correct.