Question

determine the derivative of the following function by applying the quotient rule. (h(x)=\frac{ln(x^{2}+x)}{x^{2}+1}). select the correct answer below. (h(x)=\frac{\frac{2x + 1}{x^{2}+x}(x^{2}+1)-2xln(x^{2}+x)}{(x^{2}+1)^{2}}) (h(x)=\frac{\frac{2x + 1}{x^{2}+x}(x^{2}+1)-2xln(x^{2}+x)}{x^{2}+1}) (h(x)=\frac{\frac{2x + 1}{x^{2}+1}-2xln(x^{2}+x)}{(x^{2}+1)^{2}}) (h(x)=\frac{\frac{2x + 1}{x^{2}+x}+2xln(x^{2}+x)}{(x^{2}+1)^{2}})

Explanation:

Step1: Recall quotient - rule

Quotient rule: If $h(x)=\frac{u(x)}{v(x)}$, then $h'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$. Here $u(x)=\ln(x^{2}+x)$ and $v(x)=x^{2}+1$.

Step2: Differentiate $u(x)$ and $v(x)$

$u'(x)=\frac{2x + 1}{x^{2}+x}$, $v'(x)=2x$.

Step3: Apply quotient - rule

$h'(x)=\frac{\frac{2x + 1}{x^{2}+x}(x^{2}+1)-2x\ln(x^{2}+x)}{(x^{2}+1)^{2}}$.

Answer:

The first option: $h'(x)=\frac{\frac{2x + 1}{x^{2}+x}(x^{2}+1)-2x\ln(x^{2}+x)}{(x^{2}+1)^{2}}$