Question

Differentiate y=(ln(ax+b))^(x^n).
Can anyone get me started on this "chain rule mess"?

Answer 1

what is the "outside" function?

Answer 2

say what

Answer 3

Henpen: You can't put a bracket between x and n, if you could then (7^(2^3)) would be the same as (7^2)^3, which it is not!

Answer 4

Yes, of course.

Answer 5

zzr0ck3r: I'm not entirely sure, a^x maybe, but then the inside function is x^n, and then I'm missing something

Answer 6

\[y=(\ln(ax+b))^{x^n}\]
\[\frac{d(\ln(ax+b))^{x^n} }{dln(ax+b))}\frac{dln(ax+b))}{dx}\]

Answer 7

1st term is \[x^nln(ax+b))^{x^n-1}\]or\[x^n\frac{ln(ax+b))^{x^n}}{ln(ax+b)}\]

Answer 8

You've made me wary now- I've probably made another mistake, but from here it looks sound.

Answer 9

xnln(ax+b))xn−1 , that is using the power rule on an exponential function, that won't work

Answer 10

Well, have to go to school now, guess I'll have the answer tonight

Answer 11

@TEL74 , @henpen here is the answer : first in general form , then applied to this case

Answer 12

\[{f(x)^{g(x)} }' =(g(x) - 1)*f(x)^{g(x)} * f'(x) + f(x)^{g(x)} *g'(x)\]

Answer 13

\[(x^n-1)*\ln(ax+b)^{(x^n-1)}*(ax+b)^{-1} + nx^{n-1}*\ln(ax+b)^{x^n}\]

Answer 14

@TEL74 , @henpen here is the answer

Answer 15

@TEL74, I apologise.

Mikael, what's the proof for that? I looked on wolfram and found\[\frac{df(x)^{g'(x)}}{dx}=\frac{df(x)^{g'(x)}}{df(x)}\frac{df(x)}{dx}+\frac{df(x)^{g'(x)}}{dg'(x)}\frac{dg'(x)}{dx}\]Which looks like\[\frac{df(x)^{g'(x)}}{dx}=\frac{df(x)^{g'(x)}}{dx}+\frac{df(x)^{g'(x)}}{dx}\]to me.

Answer 16

@henpen hi , it seems that the first POWER on the left is not the one in your question.
Besides that - THIS IS EXACTLY THE FPRMULA THAT LEADS TO WHAT I CALLED "THE GENERAL FORM"

Answer 17

Here is the explanation of Chain rule that I never saw in any place, so I devised it and it is the clearest I know.