if $\\frac{x^{2}}{36}+\\frac{y^{2}}{16}=1$ is the equation of an ellipse, and the point $(3,3.46)$ is on its…

Explanation:

Step1: Differentiate both sides of the equation

Differentiate (\frac{x^{2}}{36}+\frac{y^{2}}{16} = 1) with respect to (x). Using the sum rule ((u + v)^\prime=u^\prime + v^\prime), where (u=\frac{x^{2}}{36}) and (v = \frac{y^{2}}{16}). For (u=\frac{x^{2}}{36}), by the power rule ((x^{n})^\prime=nx^{n - 1}), (u^\prime=\frac{2x}{36}=\frac{x}{18}). For (v=\frac{y^{2}}{16}), using the chain - rule ((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)) (here (f(u)=\frac{u^{2}}{16}), (u = y(x))), (v^\prime=\frac{2y}{16}\cdot y^\prime=\frac{y}{8}y^\prime). The derivative of the right - hand side (since the derivative of a constant (C) is (0)) is (0). So, (\frac{x}{18}+\frac{y}{8}y^\prime=0).

Step2: Solve for (y^\prime)

Subtract (\frac{x}{18}) from both sides of the equation (\frac{x}{18}+\frac{y}{8}y^\prime=0) to get (\frac{y}{8}y^\prime=-\frac{x}{18}). Then, multiply both sides by (\frac{8}{y}) (assuming (y\neq0)) to isolate (y^\prime). So, (y^\prime=-\frac{8x}{18y}=-\frac{4x}{9y}).

Step3: Substitute (x = 3) and (y = 3.46)

Substitute (x = 3) and (y = 3.46) into the formula for (y^\prime). (y^\prime=-\frac{4\times3}{9\times3.46}). First, calculate (4\times3 = 12) and (9\times3.46=31.14). Then, (y^\prime=-\frac{12}{31.14}\approx - 0.385).

Answer:

(-0.385)