Solving Arctan Problem: y'=\frac{1}{1-(lnx)^{-2}}

[UrbanXrisis]

y=tan^{-1}\left(\frac{1}{ln(x)}\right)
y'=\frac{1}{1-\left(\frac{1}{ln(x)}\right)^2}
y'=\frac{1}{1-(lnx)^{-2}}

is this correct?

do I have to take into account the derivative of \frac{1}{ln(x)}?

if I do, what is the derivative of \frac{1}{ln(x)}?

 

[Pyrrhus]

Yes, you need to take into account the derivative of \frac{1}{\ln(x)}.

You could either do the derivative of a division or do the derivative of (\ln(x))^{-1}.

 

[UrbanXrisis]

y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}
y'=\frac{1}{x(lnx^2)-x}}

is that it?

 

[Pyrrhus]

y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}
y'=\frac{1}{x(lnx^2)-x}}

is that it?

Little changes...

y'=\frac{1}{1+(lnx)^{-2}}*\frac{-1}{x(\ln(x))^{2}}

Something else i noticed

(tan^{-1} (f(x)))' = \frac{f'(x)}{1+(f(x))^2}

 

[UrbanXrisis]

can that become y'=\frac{1}{-x(lnx^2)+x}}?

 

[Pyrrhus]

Stop putting the square next to the x like that, use parentheses.

 

[UrbanXrisis]

sorry, didn't notice that

y'=\frac{1}{-x(lnx)^2+x}}

that's what I meant

 

[Pyrrhus]

Sure, except for the sign problem.

 

[UrbanXrisis]

y'=\frac{1}{-x\ln(x)^{2}-x}