Question

the point $(-3, 4)$ is on the curve defined by $x^2y^3 = 576$. which of the following statements is true about the curve at the point $(-3, 4)$?
a $\frac{dy}{dx} < 0$ and $\frac{d^2y}{dx^2} < 0$
b $\frac{dy}{dx} < 0$ and $\frac{d^2y}{dx^2} > 0$
c $\frac{dy}{dx} > 0$ and $\frac{d^2y}{dx^2} < 0$
d $\frac{dy}{dx} > 0$ and $\frac{d^2y}{dx^2} > 0$

Explanation:

Step1: Find \(\frac{dy}{dx}\) using implicit differentiation

Given the equation \(x^{2}y^{3}=576\), differentiate both sides with respect to \(x\).
Using the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{2}\) and \(v = y^{3}\).
\(\frac{d}{dx}(x^{2}y^{3})=\frac{d}{dx}(576)\)
\(2xy^{3}+x^{2}\cdot3y^{2}\frac{dy}{dx}=0\)
Solve for \(\frac{dy}{dx}\):
\(3x^{2}y^{2}\frac{dy}{dx}=- 2xy^{3}\)
\(\frac{dy}{dx}=\frac{-2xy^{3}}{3x^{2}y^{2}}=\frac{-2y}{3x}\)
Substitute \(x=-3\) and \(y = 4\) into \(\frac{dy}{dx}\):
\(\frac{dy}{dx}=\frac{-2\times4}{3\times(-3)}=\frac{-8}{-9}=\frac{8}{9}>0\)

Step2: Find \(\frac{d^{2}y}{dx^{2}}\) using quotient rule or product rule

First, we have \(\frac{dy}{dx}=-\frac{2}{3}\cdot\frac{y}{x}\)
Differentiate \(\frac{dy}{dx}\) with respect to \(x\) using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = y\) and \(v=x\)
\(\frac{d^{2}y}{dx^{2}}=-\frac{2}{3}\cdot\frac{\frac{dy}{dx}\cdot x - y\cdot1}{x^{2}}\)
We know \(\frac{dy}{dx}=\frac{-2y}{3x}\), substitute it into the above formula:
\(\frac{d^{2}y}{dx^{2}}=-\frac{2}{3}\cdot\frac{(\frac{-2y}{3x})\cdot x - y}{x^{2}}\)
Simplify the numerator inside the fraction:
\((\frac{-2y}{3x})\cdot x - y=\frac{-2y}{3}-y=\frac{-2y - 3y}{3}=\frac{-5y}{3}\)
So \(\frac{d^{2}y}{dx^{2}}=-\frac{2}{3}\cdot\frac{\frac{-5y}{3}}{x^{2}}=-\frac{2}{3}\cdot\frac{-5y}{3x^{2}}=\frac{10y}{9x^{2}}\)
Substitute \(x = - 3\) and \(y = 4\) into \(\frac{d^{2}y}{dx^{2}}\):
\(x^{2}=(-3)^{2} = 9\), \(y = 4\)
\(\frac{d^{2}y}{dx^{2}}=\frac{10\times4}{9\times9}=\frac{40}{81}>0\)

Answer:

D. \(\frac{dy}{dx}>0\) and \(\frac{d^{2}y}{dx^{2}}>0\)