Differentiating [ sin(1/ln(x)) / x ] solution?

[r3dxP]

[SOLVED] differentiating [ sin(1/ln(x)) / x ].. solution?

hello all. I do not have the solution to this question that I am about to ask. But if you find the time, try this solving this problem and feel free to type your answer and compare with mine.
differentiate :: sin(1/ln(x)) / x

my answer: -[ ( (sin(1/lnx)*(ln(x)^2) + cos(1/ln(x)) ) / ( (x^2)*(ln(x)^2) ) ]

i am extremely curious about what you all got. I want to see if i did it correctly. Thanks in advance.

I found some time to find the d^2y/dx^2 of this f().

answer #2 for d^2y/dx^2 : { [ (cos(1/lnx))(2+4lnx+2(lnx)^2+(x^2)(lnx)^3) ]+[ (sin(1/lnx))* (lnx) * (4(lnx)^2 - x^2+2(lnx)^3) ] } / { [(lnx)*x]^3 }

woah, i spent so much time doing it.. if you guys have a time to do it, post your answer pls! thanks again.

 

[HallsofIvy]

Pretty straight forward, just a little tedius. (ln x)'= 1/x so (1/ln x)'= ((ln x)^-1)'= -(ln x)^-2(1/x)= -1/(x(ln x)²).

(sin(1/ln x))'= cos(1/ln x)(1/ln x)'= -cos(1/ln x)/(x(ln x)²).


By the quotient rule, (sin(1/ln x)/x)'= ((sin(1/ln x))'x- sin(1/ln x))/x²=
-cos(1/ln x)/(x³(ln x)²)- sin(1/ln x)/x².
I think, except for the fact that I separted the two fractions, that's the same thing you have.

By the way, this belongs in the "Calculus" section, not the "Differential Equations" section. The fact that the problem involves a derivative does not mean it is a differential equation!

 

[r3dxP]

sorry about that, if you don't mind, could you move this thread to the "Calculus" section? Thanks.

 

[Gaz031]

Besides your expression for the derivative don't forget to state x > 0 , x \neq 1 as a derivative at x_{1} needs to be defined over some interval x_{1}-R<x<x_{1}+R

 

[Orion1]

\boxed{\frac{d}{dx} \left( \frac{ \sin \left[ \frac{1}{\ln x} \right]}{x} \right) = - \frac{1}{x^2} \left( \frac{ \cos \left[ \frac{1}{\ln x} \right]}{(\ln x)^2} + \sin \left[ \frac{1}{\ln x} \right] \right)}

 

[Yegor]

First derivative is OK.
But the second isn't.
I got
\frac{d^2}{dx^2} \left( \frac{ \sin \left[ \frac{1}{\ln x} \right]}{x} \right) = \frac{ 2\cos( \frac{1}{\ln x}) - \frac{\sin( \frac{1}{\ln x})}{\ln x}+ 2(\ln x)^3 \sin ( \frac{1}{\ln x}) +3 \ln x\cos \left[ \frac{1}{\ln x} \right]) }{x^3 (\ln x)^3}
mathematica gives the same