if 5 - x² + 4x + y - 4y² = 0 then find dy/dx at the point (4, -1). answer dy/dx|(4,-1) =

Explanation:

Step1: Differentiate both sides

Differentiate $5 - x^{2}+4x + y-4y^{2}=0$ with respect to $x$. $\frac{d}{dx}(5)-\frac{d}{dx}(x^{2})+\frac{d}{dx}(4x)+\frac{d}{dx}(y)-\frac{d}{dx}(4y^{2}) = 0$. Since $\frac{d}{dx}(5)=0$, $\frac{d}{dx}(x^{2}) = 2x$, $\frac{d}{dx}(4x)=4$, and using the chain - rule $\frac{d}{dx}(4y^{2})=8y\frac{dy}{dx}$, we get $0 - 2x+4+\frac{dy}{dx}-8y\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side: $\frac{dy}{dx}-8y\frac{dy}{dx}=2x - 4$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(1 - 8y)=2x - 4$. Then $\frac{dy}{dx}=\frac{2x - 4}{1 - 8y}$.

Step3: Substitute the point

Substitute $x = 4$ and $y=-1$ into $\frac{dy}{dx}=\frac{2x - 4}{1 - 8y}$. $\frac{dy}{dx}\big|_{(4,-1)}=\frac{2\times4 - 4}{1-8\times(-1)}=\frac{8 - 4}{1 + 8}=\frac{4}{9}$.

Answer:

$\frac{4}{9}$