homework5: problem 32 (1 point) suppose that $f(x)=\frac{4}{ln(x^{2}+3)}$. find $f(1)$. $f(1)=$ you have…

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4$, so $u'=0$, and $v=\ln(x^{2}+3)$, so $v'=\frac{2x}{x^{2}+3}$ by the chain - rule. Then $f'(x)=\frac{0\times\ln(x^{2}+3)-4\times\frac{2x}{x^{2}+3}}{(\ln(x^{2}+3))^{2}}=-\frac{8x}{(x^{2}+3)(\ln(x^{2}+3))^{2}}$.

Step2: Substitute $x = 1$

Substitute $x = 1$ into $f'(x)$. When $x = 1$, we have $x^{2}+3=1 + 3=4$ and $\ln(x^{2}+3)=\ln4$. Then $f'(1)=-\frac{8\times1}{4\times(\ln4)^{2}}=-\frac{2}{(\ln4)^{2}}$.

Answer:

$-\frac{2}{(\ln4)^{2}}$