3. if $x^{2}+y^{2}=25$, what is the value of $\frac{d^{2} y}{d x^{2}}$ at the point $(4,3)$? (a)…

Explanation:

Step1: Differentiate the equation (x^{2}+y^{2}=25) with respect to (x)

Using implicit differentiation, we have (2x + 2y\frac{dy}{dx}=0). Simplify to get (\frac{dy}{dx}=-\frac{x}{y}).

Step2: Differentiate (\frac{dy}{dx}=-\frac{x}{y}) with respect to (x)

Using the quotient rule (\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}), where (u=-x) and (v = y). (\frac{d^{2}y}{dx^{2}}=-\frac{y - x\frac{dy}{dx}}{y^{2}}). Substitute (\frac{dy}{dx}=-\frac{x}{y}) into the above formula: (\frac{d^{2}y}{dx^{2}}=-\frac{y - x\left(-\frac{x}{y}\right)}{y^{2}}=-\frac{y^{2}+x^{2}}{y^{3}}).

Step3: Substitute (x = 4) and (y = 3) into (\frac{d^{2}y}{dx^{2}}=-\frac{y^{2}+x^{2}}{y^{3}})

Since (x^{2}+y^{2}=25), then (\frac{d^{2}y}{dx^{2}}=-\frac{25}{3^{3}}=-\frac{25}{27}).

Answer:

A. (-\frac{25}{27})