Solving tricky derivative help!

[Sota Vik]

I have been trying to solve this derivative for a couple hours now to no avail! Any help would be appreciated.

 

[willyengland]

What you always can try is:

derivative of (ln(x))^sec(x) - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

 

[Deleted member 4993]

View attachment 22453
I have been trying to solve this derivative for a couple hours now to no avail! Any help would be appreciated.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Sota Vik]

What you always can try is:

derivative of (ln(x))^sec(x) - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

Unfortunately it gives me the wrong answer

 

[Deleted member 4993]

Unfortunately it gives me the wrong answer

Does that mean that you know the answer?

Can you please share the answer given to you?

 

[HallsofIvy]

$\displaystyle y= (ln(x))^{sec(x)}$ so $\displaystyle ln(y)= sec(x)ln(ln(x))$.

Differentiating that, $\displaystyle \frac{1}{y}\frac{dy}{dx}= [sec(x)tan(x)]ln(ln(x))+ sec(x)\frac{1}{ln(x)}\frac{1}{x}$....................corrected

So $\displaystyle \frac{dy}{dx}= y([sec(x)tan(x)]ln(ln(x))+ sec(x)\frac{1}{(ln(x))}\frac{1}{x})$ ....................corrected

$\displaystyle = (ln(x))^{sec(x)}([sec(x)tan(x)]ln(ln(x))+ sec(x)\frac{1}{ln((x))}\frac{1}{x})$.....................corrected

 

[Steven G]

To OP, Prof Halls made a slight mistake. Starting with the 2nd line where ever you see ln(ln(x)) it should be ln(x).

 

[Deleted member 4993]

To OP, Prof Halls made a slight mistake. Starting with the 2nd line where ever you see ln(ln(x)) it should be ln(x).

Please write (post) the corrected version you propose.

 

[Cubist]

To OP, Prof Halls made a slight mistake. Starting with the 2nd line where ever you see ln(ln(x)) it should be ln(x).

I half agree with Jomo

Differentiating that, $\displaystyle \frac{1}{y}\frac{dy}{dx}= [sec(x)tan(x)]ln(ln(x))+ sec(x)\frac{1}{ln(ln(x))}\frac{1}{x}$

HallsofIvy got it mostly correct but I think there's a slight mistake (corrected now)shown by the red highlight below (note the first ln(ln(x)) is correct)...

$\displaystyle \frac{1}{y}\frac{dy}{dx}= [sec(x)tan(x)]ln(ln(x))+ sec(x)\frac{1}{\color{red}ln(ln(x))\color{black}}\frac{1}{x}$

So this line ought to be...

$\displaystyle \frac{1}{y}\frac{dy}{dx}= [\sec(x)\tan(x)]\ln(\ln(x))+ \sec(x)\frac{1}{\ln(x)}\frac{1}{x}$

Continuing to the final answer, simplify

$\displaystyle = \sec(x)\left(\tan(x)\ln(\ln(x))+ \frac{1}{x\ln(x)}\right)$

Multiply both sides by y

$\displaystyle \frac{dy}{dx} =\left[ \ln(x)\right]^{sec(x)}\sec(x)\left(\tan(x)\ln(\ln(x))+ \frac{1}{x\ln(x)}\right)$

Wolfram drew a graph that looks correct, but their symbolic result seems incorrect because I plotted their result and it didn't match their own graph (but maybe I made a typo)