5. use implicit differentiation to find y. for 2xy + y² = 5

Explanation:

Step1: Differentiate both sides

Differentiate $2xy + y^{2}=5$ with respect to $x$. Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ for $2xy$ where $u = 2x$ and $v = y$, and the chain - rule for $y^{2}$. $(2xy)^\prime+(y^{2})^\prime=(5)^\prime$. $2y + 2xy^\prime+2yy^\prime = 0$.

Step2: Solve for $y^\prime$

Factor out $y^\prime$: $2xy^\prime+2yy^\prime=-2y$. $y^\prime(2x + 2y)=-2y$. $y^\prime=\frac{-2y}{2x + 2y}=\frac{-y}{x + y}$.

Step3: Differentiate $y^\prime$ to find $y^{\prime\prime}$

Using the quotient - rule $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$ where $u=-y$ and $v=x + y$. $u^\prime=-y^\prime$ and $v^\prime=1 + y^\prime$. $y^{\prime\prime}=\frac{-y^\prime(x + y)-(-y)(1 + y^\prime)}{(x + y)^{2}}$. Substitute $y^\prime=\frac{-y}{x + y}$ into the above formula: [ \begin{align*} y^{\prime\prime}&=\frac{-\frac{-y}{x + y}(x + y)+y(1-\frac{y}{x + y})}{(x + y)^{2}}\ &=\frac{y + y(\frac{x + y - y}{x + y})}{(x + y)^{2}}\ &=\frac{y+\frac{xy}{x + y}}{(x + y)^{2}}\ &=\frac{\frac{y(x + y)+xy}{x + y}}{(x + y)^{2}}\ &=\frac{xy+y^{2}+xy}{(x + y)^{3}}\ &=\frac{2xy + y^{2}}{(x + y)^{3}} \end{align*} ] Since $2xy + y^{2}=5$, then $y^{\prime\prime}=\frac{5}{(x + y)^{3}}$.

Answer:

$y^{\prime\prime}=\frac{5}{(x + y)^{3}}$