Is the derivative for any equation ln (kx) for any positive constant k always 1/x ?

Answer 1

Yes. If

f(x) = ln(kx), then

f'(x) = 1/(kx) {k}
f'(x) = k/(kx)
f'(x) = 1/x

The reason why this is so is because of the log properties. Note that if

f(x) = ln(kx), we can split this apart into two.

f(x) = ln(k) + ln(x)

Since ln(k) is a constant, its derivative is 0, and

f'(x) = 0 + 1/x
f'(x) = 1/x

Answer 2

Yes... with the caveat that ln(x) is only defined for x greater then 0. To do the math simply substitute u for kx and remember the derivative of ln(x) is 1/x.

Answer 3

surely. imperative (a million/x dx) = ln|x| + C yet C is the comparable as ln(e^c), as ln and e are inverses of one yet another, meaning we get ln|x| + ln(e^c) Combining those 2 utilising the log sources that log[base b](a) + log[base b](c) = log[base b](ac), we get ln| e^c (x) | yet e^c is in simple terms a relentless obtained from a relentless, so why not in simple terms call it a relentless, ok? ln|kx| is our answer.

Answer 4

yes.

(d/dx) ln u = u'/u

ln kx = k/kx = 1/x