To differentiate the function $g(x) = \ln\left( \frac{7 - x}{7 + x} \right)$, we can use the chain rule and the derivative properties of the natural logarithm.

Steps:

1. Let $u = \frac{7 - x}{7 + x}$.
2. Then, $g(x) = \ln(u)$.
3. The derivative of $g(x)$ with respect to $x$ is given by $g'(x) = \frac{1}{u} \cdot \frac{du}{dx}$.

Now we need to find $\frac{du}{dx}$.

Finding $\frac{du}{dx}$:

$u = \frac{7 - x}{7 + x}$ We can use the quotient rule, where if $u = \frac{f(x)}{g(x)}$, then: $\frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

Here, $f(x) = 7 - x$ and $g(x) = 7 + x$.

• $f'(x) = -1$
• $g'(x) = 1$

Applying the quotient rule: $\frac{du}{dx} = \frac{(-1)(7 + x) - (7 - x)(1)}{(7 + x)^2} = \frac{-7 - x - 7 + x}{(7 + x)^2} = \frac{-14}{(7 + x)^2}$

Now, substituting back:

$g'(x) = \frac{1}{\frac{7 - x}{7 + x}} \cdot \frac{-14}{(7 + x)^2}$

Simplify further: $g'(x) = \frac{7 + x}{7 - x} \cdot \frac{-14}{(7 + x)^2}$

$g'(x) = \frac{-14}{(7 - x)(7 + x)}$

Finally, we get: $g'(x) = \frac{-14}{49 - x^2}$

So, the derivative $g'(x) = \frac{-14}{49 - x^2}$.

Do you want more details or have any questions?

Related Questions:

1. How do you find the derivative of a logarithmic function involving a different base?
2. What is the chain rule in calculus, and when is it used?
3. How can we differentiate a function involving trigonometric expressions?
4. What are the common properties of logarithms useful in differentiation?
5. How would you integrate the function $g'(x)$ to find the original $g(x)$?

Tip: Always simplify the argument inside the logarithm before differentiating, if possible, to make the differentiation process easier.