Question

Need guidance in differentiating f(x)=arctan(√(ln(1/x))

Answer 1

\[\arctan(\sqrt{\ln(\frac{ 1 }{ x }})\]

Answer 2

\[\frac{ d }{ dx } (\tan^{-1}x)=\frac{ 1 }{ 1+x^2 }\] so you will have to use chain rule here, is your derivative \[ \frac{ d }{ dx } \arctan \left( \sqrt{\ln (\frac{ 1 }{ x })}\right)\]

Answer 3

I think it might be easier if you like \[\sqrt{\ln \frac{ 1 }{ x }} = u\] and then make the substitution back so you can see what exactly is going on

Answer 4

let*

Answer 5

Or you can just take the derivative haha, it's going to be ugly, try using the chain rule and see what you get

Answer 6

so I would get \[\frac{ 1 }{ \ln \frac{ 1 }{ x } } x \frac{ d }{ dx }\sqrt{\ln \frac{ 1 }{ x }}\], then the chain rule again?

Answer 7

sorry, 1/(1+ln(1/x))

Answer 8

\[\frac{ d }{ dx } \arctan \left( \sqrt{\ln (\frac{ 1 }{ x })}\right) = \frac{ 1 }{ 1+\ln(1/x) } * \ln(\frac{ 1 }{ x })'\] do you mean this

Answer 9

Yeah. Im not really fimiliar with the tools yet, haha

Answer 10

Then since its the derivative of ln(1/x), we would have to apply the chain rule again?

Answer 11

One sec it's not as easy as that, let me try on paper and I'll get back to you in a second

Answer 12

Ok so we should get \[- \frac{ 1 }{ \sqrt{\ln \left( \frac{ 1 }{ x } \right)}(2x+2xln \left( \frac{ 1 }{ x } \right) }\] it's probably easier making the u substitution and breaking it in parts, or we can do it all at once which is kind of nasty

Answer 13

\[\frac{ d }{ dx } \left( \sqrt{\ln(\frac{ 1 }{ x })} \right)\] so what's the derivative of this?

Answer 14

i believe it would be \[\frac{ -1 }{ 2x \sqrt{\ln \frac{ 1 }{ x }} }\]

Answer 15

Right, awesome!

Answer 16

Ok now we have \[\frac{ d }{ dx } arctanu= \frac{ 1 }{ 1+u^2 }\] so now we just take \[u = \sqrt{\ln(1/x)}\] and substitute that in, and we can now multiply by that derivative above and we're done :)

Answer 17

Does that make sense?

Answer 18

kind of, do we not need to use the chain rule again for 1/x?

Answer 19

Only reason I broke it up is because we would be hesitant in taking the derivative of just ln(1/x) which is wrong

Answer 20

We did take the derivative, we took the derivative of \[\sqrt{\ln(1/x)}\] which included that

Answer 21

Ok I guess I'll show the long ugly way to, \[\frac{ d }{ dx } \arctan(\sqrt{\ln(1/x)}) = \frac{ 1 }{ 1+\left( \sqrt{\ln(1/x)} \right)^2 } \times [\sqrt{\ln(1/x)}]'\] I guess it's not too bad but you can't square for the derivative

Answer 22

OH, duh so we get down to \[\frac{ 1 }{ 1+\ln \frac{ 1 }{ x } }*\frac{ -1 }{ 2x \sqrt{\ln \frac{ 1 }{ x }} }\]

Answer 23

Yes, exactly :) so we then have \[-\frac{ 1 }{ 2xln \sqrt{1/x}(1+\ln(1/x)) }\]

Answer 24

Is this the same as the one you had above?

Answer 25

Yup, it doesn't matter

Answer 26

Oh, okay awesome. Thank you for your help!

Answer 27

The above just multiplied through by 2x