Question

differentiate y = ln(x)^ln(x)

Answer 1

So I used logarithmic differentiation.

Answer 2

From there I got to dy/dx =y(ln(ln(x))/x+1/x) but apparently its incorrect

Answer 3

take log from both sides

Answer 4

yes i did that

Answer 5

so you get logy=xln(x), now try differentiating

Answer 6

dy/y=(ln(x)+1)dx, now dy/dx=y(ln(x)+1)

Answer 7

I thought it worked like this ln(y) = ln(x) *ln((ln(x))

Answer 8

I will derive the general formula.
If you want you can do these derivations
yourself, they are pretty straight forward.

\(\color{#000000}{ \displaystyle y=f(x)^{g(x)} }\)

Take the natural log of both sides:
\(\color{#000000}{ \displaystyle \ln y=\ln \left[ f(x)^{g(x)}\right] }\)
\(\color{#000000}{ \displaystyle \ln y=g(x)\ln \left[ f(x)\right] }\)

Differentiate using product rule, and chain rule:
\(\color{#000000}{ \displaystyle \frac{y'}{y} =g'(x)\ln \left[ f(x)\right] +g(x)\frac{f'(x)}{ f(x)} }\)

\(\color{#000000}{ \displaystyle y' =y\left[g'(x)\ln \left[ f(x)\right] +g(x)\frac{f'(x)}{ f(x)} \right] }\)

Plug the y you had in the beginning:
\(\color{#000000}{ \displaystyle y' =f(x)^{g(x)} \left[g'(x)\ln \left[ f(x)\right] +g(x)\frac{f'(x)}{ f(x)} \right] }\)

Answer 9

and therefore, dy/dx=ln(x)^ln(x)(ln(x)+1)

Answer 10

if anything you can use wolfram to check your work.

Answer 11

Oh apparently the last step @SolomonZelman wrote about plugging y back in was the part I was missing. Thanks everyone for the help

Answer 12

Oh, lol. Well, now you know:)
Good luck.