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]

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