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.
The power rule for finding the derivative of ln(u)^k is k*(ln(u))^(k-1) * (u')/u.
Yes, the chain rule can be used to find the derivative of ln(u)^k. The derivative can be rewritten as k*ln(u)^(k-1) * (u')/u, where ln(u)^(k-1) is the outer function and u is the inner function.
No, the derivative of ln(u)^k is not the same as k*ln(u). The correct derivative is k*ln(u)^(k-1) * (u')/u.
If k is a constant, the derivative of ln(u)^k is k/u.
Yes, the quotient rule can be used to find the derivative of ln(u)^k. The derivative can be rewritten as k*ln(u)^(k-1) * (u')/u, where k*ln(u)^(k-1) is the numerator and u is the denominator.