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

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SUMMARY

The discussion focuses on the differentiation of the function \( y = \tan^{-1}\left(\frac{1}{\ln(x)}\right) \) and its derivative \( y' = \frac{1}{1 - (\ln x)^{-2}} \). Participants confirm that the derivative of \( \frac{1}{\ln(x)} \) must be considered, leading to the expression \( y' = \frac{1}{1 - (\ln x)^{-2}} \cdot \frac{-1}{x(\ln(x))^2} \). The final corrected form of the derivative is \( y' = \frac{1}{-x(\ln(x))^2 - x} \), emphasizing the importance of proper notation and sign conventions in calculus.

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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)}?
 
Last edited:
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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}.
 
y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}
y'=\frac{1}{x(lnx^2)-x}}

is that it?
 
UrbanXrisis said:
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}
 
Last edited:
can that become y'=\frac{1}{-x(lnx^2)+x}}?
 
Stop putting the square next to the x like that, use parentheses.
 
sorry, didn't notice that

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

that's what I meant
 
Sure, except for the sign problem.
 
y'=\frac{1}{-x\ln(x)^{2}-x}
 

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