Calculus help? Find y': if y= (ln x) ^ sin x

Find y': if y= (ln x) ^ sin x

Please can anyone help? I have posted this before and one person ignored the ( ^ ) sign. Another person doesn't think that (^ sin x) can be possible and that sin x cannot be an exponent. This is a REAL question. If anyone can explain how to do this without diasgreeing or ignoring parts of it can you please contribute and explain it a step by step process so that I can understand?

Thanks!

Logarithmic and implicit differentiation:

y = ln(x)^sin(x)
ln(y) = ln(ln(x)^sin(x))
ln(y) = sin(x) * ln(ln(x))
dy / y = sin(x) * (1/ln(x)) * (1/x) * dx + ln(ln(x)) * cos(x) * dx
dy / y = dx * (sin(x) / (x * ln(x)) + ln(ln(x)) * cos(x))
dy/dx = y * (sin(x) / (x * ln(x)) + ln(ln(x)) * cos(x))
dy/dx = (ln(x)^sin(x)) * (sin(x) / (x * ln(x)) + ln(ln(x)) * cos(x))

It is a variable to the power of a variable so you use the exponential rule, which is taking the natural log of both sides so...

lny = ln(lnx)^sinx

Using properties of logs...

lny = sinx[ln(lnx)]

Now take the derivative of BOTH SIDES... Also need to use chain rule:

1/y * dy/dx = sinx * (1/lnx * 1/x) + ln(lnx) * cosx

Simplify.

dy/dx = y[(sinx/xlnx) + ln(lnx)cosx]

Substitute y back into the equation...

dy/dx = (lnx)^sinx[(sinx/xlnx) + ln(lnx)cosx]

I like to approach these sorts of problems in a more general way.
Maybe this will help you - if not, ignore it.
Let f(x)=g(x)^h(x) so that
ln[f(x)]=h(x)ln[g(x)]. Differentiating this yields
f'(x)/f(x)=h(x)*g'(x)/g(x)+h'(x)ln[g(x… and so

f'(x)=f(x){ h(x)*g'(x)/g(x)+h'(x)ln[g(x)] }. (*)
In your problem
g(x)=ln(x) and h(x)=sin(x) g'(x)=1/[xln(x)] and h'(x)=cos(x).
Now just put everything together using (*).

y = lnx ^ sinx
ln y = sinx ln (ln x)
1 / y * dy / dx = (sinx/(x ln x)) + cos x (ln(ln x))
dy / dx = y [(sinx/(x ln x)) + cos x (ln(ln x))]
dy / dx = (ln x^sin x)[(sinx/(x ln x)) + cos x (ln(ln x))]

dk