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