The discussion revolves around finding the derivative of the function f(x) = ln(u)^k, where u is a function of x and k is a constant. Participants explore different interpretations of the expression and the implications for differentiation, focusing on the application of logarithmic differentiation and simplification techniques.
There is disagreement regarding the interpretation of the function f(x) and the correct approach to finding its derivative. Some participants believe the exponent applies to ln(u), while others argue it applies to u, leading to different derivative formulations. The discussion remains unresolved.
Participants express uncertainty about the notation used in the original function and how it affects the differentiation process. There is also a mention of previous discussions that may influence the current interpretations.
No, it's not. The derivative does not involve a log.silvershine said:This is a concept that confuses me. If you have a function where f(x) = ln(u)^k (where u is a function and k is an unknown constant), then how would you solve for the derivative? I've already guessed it would be
f'(x) = [kln(u)^(k-1)](1/u)(u')
Is that right?
Mark44 said:No, it's not. The derivative does not involve a log.
If f(x) = ln[uk], where u is a differentiable function of x, then
$$f'(x) = \frac{1}{u^k}\cdot ku^{k-1}\cdot u'$$
$$= \frac{k}{u}\cdot u'$$
This is essentially the same problem that you asked in the other thread you posted. As I said there, you can do things this way, but it is much simpler to simplify the log expression before you differentiate.
If we simplify first, we have
f(x) = k ln(u)
So f'(x) = k * d/dx(ln(u)) = k * 1/u * u' = (k/u)* u', which is the same as I got by differentiating without simplifying first.
For readers who didn't see the other thread, I'm assuming that what you wrote as ln(u)^k means ln[uk], and not [ln(u)]k.