Question

1. if $f(x)=ln|x^{2}-1|$, then $f(x)=$

Explanation:

Step1: Apply the chain rule

Let \(u = x^{2}-1\), then \(y=\ln|u|\). The derivative of \(\ln|u|\) with respect to \(u\) is \(\frac{1}{u}\), and the derivative of \(u = x^{2}-1\) with respect to \(x\) is \(2x\).
By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
So \(\frac{dy}{dx}=\frac{1}{u}\cdot2x\).

Step2: Substitute \(u\) back

Since \(u = x^{2}-1\), substituting \(u\) into the expression from Step 1 gives \(\frac{2x}{x^{2}-1}\).

Answer:

\(\frac{2x}{x^{2}-1}\)