What is the derivative of ln(u)^k?

[silvershine]

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?

 

[Robert1986]

looks right to me

 

[Mark44]

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?

No, it's not. The derivative does not involve a log.

If f(x) = ln[u^k], 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[u^k], and not [ln(u)]^k.

 

[Robert1986]

No, it's not. The derivative does not involve a log.

If f(x) = ln[u^k], 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[u^k], and not [ln(u)]^k.

Ha, yes, I misread the OP - I thought the exponent was on the lnu. My mistake.

 

[Same-same]

I think the power is on the ln(u), not on u itself. He closed the parenthesis before adding the power, so it appears to be :
f(x) =(ln(u))^k.