Why do the derivations of $\ln(\frac{a}{b})$ and $\ln(a) - \ln(b)$ yield different results

Question

So I was doing some exercises and I noticed that for one example which was of the form $$\ln( \frac{a}{b} )$$ (with a and b being some term with x), that I was getting a different result for taking the derivation when using the logarithmic rule $$\ln( \frac{a}{b} ) = \ln(a) - \ln(b)$$ before deriving versus applying chain and quotient rules right away.

I tried this with some other examples of that form too and I always ended up getting different results, but I have no idea what would cause this to happen.

One of the examples I tried would be $$\ln( \frac{4+x}{4-x} )$$ which yields $$\frac{-8+8x}{(4+x)^3}$$ when applying chain and quotient rules right away and $$\frac{8}{16-x^2}$$ when using $$\ln( \frac{a}{b} ) = \ln(a) - \ln(b)$$ before deriving.

Would really appreciate some help, thanks in advance.

Answer 1

If you differentiate $\log\left(\frac{4+x}{4-x}\right)$ directly, what you get is$$\frac{\left(\frac{4+x}{4-x}\right)'}{\frac{4+x}{4-x}}=\frac{\frac{4-x+4-x}{(4-x)^2}}{\frac{4+x}{4-x}}=\frac8{(4-x)(4+x)}.\tag1$$And if you differentiate $\log(4+x)-\log(4-x)$, what you get is$$\frac1{4+x}+\frac1{4-x},$$which is equal to $(1)$.

Answer 2

The derivative is $\frac{8}{16-x^2}$ no matter how you solve it.

Directly (as a quotient): $$\frac{\frac{1}{(4-x)^2}}{\frac{4+x}{4-x}}(4-x+4+x) = \frac{8}{16-x^2}$$

Using the log rule ($\log(4+x) - \log(4-x)$): $$\frac{1}{4+x} + \frac{1}{4-x} = \frac{8}{16-x^2}$$