find the derivative of $h(x)=\\ln(x^{2}-3x)$.\n1 $h(x)=\\frac{1}{2x - 3}$\n2 $h(x)=\\frac{1}{x^{2}-3x}$
find the derivative of $h(x)=\\ln(x^{2}-3x)$.\n1 $h(x)=\\frac{1}{2x - 3}$\n2 $h(x)=\\frac{1}{x^{2}-3x}$
Answer
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})