if (x^{2}+y^{2}=25), what is the value of (\frac{dy}{dx}) at the point ((4,3))?\na (-\frac{4}{3})\nb…

if (x^{2}+y^{2}=25), what is the value of (\frac{dy}{dx}) at the point ((4,3))?\na (-\frac{4}{3})\nb (-\frac{3}{4})\nc (\frac{3}{4})\nd (\frac{4}{3})\ne (\frac{25}{2})
Answer
Explanation:
Step1: Differentiate implicitly
Differentiate $x^{2}+y^{2}=25$ with respect to $x$. Using the power - rule and chain - rule, we get $2x + 2y\frac{dy}{dx}=0$.
Step2: Solve for $\frac{dy}{dx}$
Isolate $\frac{dy}{dx}$ in the equation $2x + 2y\frac{dy}{dx}=0$. First, subtract $2x$ from both sides: $2y\frac{dy}{dx}=-2x$. Then divide both sides by $2y$ (assuming $y\neq0$), so $\frac{dy}{dx}=-\frac{x}{y}$.
Step3: Substitute the point $(4,3)$
Substitute $x = 4$ and $y = 3$ into $\frac{dy}{dx}=-\frac{x}{y}$. We have $\frac{dy}{dx}=-\frac{4}{3}$.
Answer:
A. $-\frac{4}{3}$