91Ó°ÊÓ

The chain rule is a fundamental tool in calculus for differentiating compositions of functions. When you encounter a function like \(y = \ln \left|\frac{1+x}{1-x}\right|\), it's clear that there's an outer function \(\ln|u|\) and an inner function \(u = \frac{1+x}{1-x}\).

The chain rule states that to differentiate a composite function, \(y\), one should multiply the derivative of the outer function by the derivative of the inner function. This can be written as:

• \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)

When applying the chain rule, always identify your inner function first, differentiate it with respect to \(x\), and then differentiate the outer function with respect to the inner function. This process helps you to systematically break down complex problems into simpler parts.

The quotient rule helps differentiate functions of the form \(\frac{f(x)}{g(x)}\). It's necessary when dealing with ratios of functions, like our inner function \(u = \frac{1+x}{1-x}\).

To apply the quotient rule, use the formula:

• \(\frac{du}{dx} = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\)

Here, \(f(x) = 1+x\) and \(g(x) = 1-x\). Differentiate each:

• \(f'(x) = 1\)
• \(g'(x) = -1\)

Substitute these derivatives back into the quotient rule formula:

• \(\frac{du}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2}\)
• This simplifies to \(\frac{2}{(1-x)^2}\)

Whenever you have to apply the quotient rule, remember to carefully find the derivative of the top and bottom functions, then substitute them back, and simplify if possible.

Differentiating the natural logarithm, \(\ln(x)\), involves understanding how changes in \(x\) affect the rate of change of its natural log. The derivative of a natural log, \(\ln(u)\), with respect to \(x\) is \(\frac{1}{u}\cdot\frac{du}{dx}\).

In our given function, we've nested the \(\ln\) function around \(\frac{1+x}{1-x}\).

To find the derivative:

• First, treat the entire rational expression as \(u\), giving us \(\frac{d}{dx}(\ln|u|) = \frac{1}{u}\cdot\frac{du}{dx}\)
• From the previous step, we know \(\frac{du}{dx} = \frac{2}{(1-x)^2}\)
• Substitute \(u = \frac{1+x}{1-x}\) and \(\frac{du}{dx}\) into the expression
• The final result is \(\frac{2}{1-x^2}\)

Natural logarithm differentiation is particularly useful in simplifying derivative expressions when the function inside is a product, quotient, or power, as shown in this example.