Answer and Step-by-step explanation:
Here's a step-by-step explanation for both parts of the question:
1) To find $\frac{dy}{dx}$ for $y=x^2 \ln(\frac{x}{x+3})$, we can use the product rule and the chain rule. The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. The chain rule states that if $f(x) = g(h(x))$, then $f'(x) = g'(h(x))h'(x)$.
Let $u(x) = x^2$ and $v(x) = \ln(\frac{x}{x+3})$. Then, $u'(x) = 2x$ and $v'(x) = \frac{d}{dx} \ln(\frac{x}{x+3})$. To find $v'(x)$, we can use the chain rule. Let $g(x) = \ln(x)$ and $h(x) = \frac{x}{x+3}$. Then, $g'(x) = \frac{1}{x}$ and $h'(x) = \frac{d}{dx} (\frac{x}{x+3})$. Using the quotient rule, we find that $h'(x) = \frac{(x+3)(1)-x(1)}{(x+3)^2} = \frac{3}{(x+3)^2}$.
Substituting these values into the chain rule formula, we get $v'(x) = g'(h(x))h'(x) = \frac{1}{\frac{x}{x+3}}\cdot\frac{3}{(x+3)^2} = \frac{3}{x(x+3)}$. Substituting these values into the product rule formula, we get $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 2x\ln(\frac{x}{x+3}) + x^2\cdot\frac{3}{x(x+3)} = 2x\ln(\frac{x}{x+3}) + \frac{3}{(x+3)}$.
2) To use logarithmic differentiation to determine $\frac{dy}{dx}$ for $y=\frac{x}{\sqrt{x^2-4}}$, we can take the natural logarithm of both sides of the equation to get $\ln(y)=\ln(\frac{x}{\sqrt{x^2-4}})=\ln(x)-\ln(\sqrt{x^2-4})=\ln(x)-\frac{1}{2}\ln(x^2-4)$.
Next, we differentiate both sides with respect to x using implicit differentiation. This gives us $\frac{1}{y}\cdot\frac{dy}{dx}=\frac{1}{x}-\frac{1}{2}\cdot\frac{2x}{(x^2-4)}=\frac{1}{x}-\frac{x}{(x^2-4)}$. Solving for $\frac{dy}{dx}$, we get $\frac{dy}{dx}=y\left(\frac{1}{x}-\frac{x}{(x^2-4)}\right)=\left(\frac{x}{\sqrt{x^2-4}}\right)\left(\frac{1}{x}-\frac{x}{(x^2-4)}\right)=\left(\sqrt{\left(\dfrac{x^2-4-x^{2}}{( x^{ 2 } - 4 )}\right)}\right)= - \dfrac { 2 } { \sqrt { x ^ { 2 } - 4 } }$.