Question

find the derivative of $h(x)=\ln(x^{2}-3x)$.
1 $h(x)=\frac{1}{2x - 3}$
2 $h(x)=\frac{1}{x^{2}-3x}$

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). For \(h(x)=\ln(u)\) where \(u = x^{2}-3x\), the derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\), and the derivative of \(u=x^{2}-3x\) with respect to \(x\) is \(u^\prime = 2x - 3\). So \(h^\prime(x)=\frac{1}{u}\cdot(2x - 3)\)

Step2: Substitute \(u=x^{2}-3x\) back in

Substitute \(u=x^{2}-3x\) into the expression from Step 1. We get \(h^\prime(x)=\frac{2x - 3}{x^{2}-3x}\)

Answer:

None of the given options (Option 1: \(h^\prime(x)=\frac{1}{2x - 3}\) and Option 2: \(h^\prime(x)=\frac{1}{x^{2}-3x}\)) are correct. The correct derivative is \(h^\prime(x)=\frac{2x - 3}{x^{2}-3x}\)