derivative of log(y) function.

Question

I am trying to find derivative of log function.

How would I apply the chain rule on a basic example?

$$\ln(y)=m+bx$$

I think it is

$$\frac{dy}{dx}=e^{m+bx} *b.$$ Is this correct?

Answer 1

$\ln y = m + bx; \tag 1$

differentiate with respect to $x$:

$\dfrac{y'}{y} = b; \tag 2$

$y' = by = be^{m + bx}, \tag 3$

since from (1)

$y = e^{m + bx}. \tag 4$

Our OP thesmallprint's result is thus correct.

Answer 2

Using the definition of a derivative, $$\frac{d}{dy}{\log y}=\lim_{\epsilon\to 0}\frac{\ln(1+\frac{\epsilon}{y})}{\epsilon}.$$ Since logarithms are monotonic, the derivative doesn't vanish at $0$, so a Taylor-series argument implies a constant $k$ exists for which $\frac{d}{dy}\log y=\frac{k}{y}$. The value of $k$ depends on the logarithm's base. One definition of $e$ is as the basis obtaining the $k=1$ natural logarithm $\ln y$.