if (2x^{2}+2x + xy = 5) and (y(5)=-11), find (y(5)) by implicit differentiation. (y(5)=square)

Explanation:

Step1: Differentiate both sides

Differentiate $2x^{2}+2x + xy$ and $5$ with respect to $x$. Using the sum - rule and product - rule, we have: $\frac{d}{dx}(2x^{2})+\frac{d}{dx}(2x)+\frac{d}{dx}(xy)=\frac{d}{dx}(5)$. $4x + 2+(y+xy')=0$.

Step2: Solve for $y'$

Rearrange the equation $4x + 2+y+xy'=0$ to isolate $y'$. $xy'=-4x - 2 - y$. $y'=\frac{-4x - 2 - y}{x}$.

Step3: Substitute $x = 5$ and $y(5)=-11$

Substitute $x = 5$ and $y=-11$ into the formula for $y'$. $y'(5)=\frac{-4\times5-2-(-11)}{5}$. $y'(5)=\frac{-20 - 2 + 11}{5}=\frac{-11}{5}$.

Answer:

$-\frac{11}{5}$