How to differentiate $g(x)=\ln\left( \frac{3 -x}{3 +x}\right)$?

Question

Differentiate $g(x)=\ln \frac{3 -x}{3 +x}$.

I tried $\ln(3-x) - \ln(3+x)$

$1/(3-x) - 1/(3+x)$

or

$1/{((3-x)/(3+x)} * (((3-x)(1) - (3+x)(-1))/(3-x)^2)$

Both attempts are wrong?

Answer 1

So $\displaystyle g(x) = \ln\left(\frac{3-x}{3+x}\right)$.

This is equal to $\ln(3-x)-\ln(3+x)$.

Applying CHAIN RULE to the first term gives us

$g'(x) = \displaystyle \frac{-1}{3-x}-\frac{1}{3+x}$.

Answer 2

$$\dfrac{d(\ln(a-x))}{dx}=\dfrac{d(\ln(a-x))}{d(a-x)}\cdot\dfrac{{d(a-x)}}{dx}=\dfrac{-1}{a-x}$$

$$\dfrac{d(\ln(a+x))}{dx}=?$$

Answer 3

$$g(x)=\ln(3-x)-\ln(3+x)$$ $$g'(x)=\frac{1}{3-x} \frac{d}{dx}(3-x)-\frac{1}{3+x} \frac{d}{dx}(3+x)$$

Answer 4

We have the standard integral:

$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C=\frac{1}{2a}\ln|\frac{-(a-x)}{a+x}|+C=\frac{1}{2a}\ln|\frac{a-x}{a+x}|+C$

Differentiating both sides,

$\frac{1}{x^2-a^2}=\frac{1}{2a}\frac{d}{dx}(\ln|\frac{a-x}{a+x}|)+0$

$\frac{d}{dx}(\ln|\frac{a-x}{a+x}|)=\frac{2a}{x^2-a^2}$

$\frac{d}{dx}(\ln|\frac{3-x}{3+x}|)=\frac{6}{x^2-9}$