given $x^{2}-xy^{2}=4y$,\na) $\frac{dy}{dx}=$\nb) write the equations of the tangent lines to the curve at…

Explanation:

Step1: Differentiate both sides implicitly

Differentiate $x^{2}-xy^{2}=4y$ with respect to $x$. For $x^{2}$, the derivative is $2x$. For $-xy^{2}$, use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = - x$ and $v = y^{2}$. So $(-xy^{2})^\prime=-y^{2}-2xy\frac{dy}{dx}$. For $4y$, the derivative is $4\frac{dy}{dx}$. We get $2x - y^{2}-2xy\frac{dy}{dx}=4\frac{dy}{dx}$.

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

Rearrange the terms to isolate $\frac{dy}{dx}$: $2x - y^{2}=4\frac{dy}{dx}+2xy\frac{dy}{dx}$. Factor out $\frac{dy}{dx}$ on the right - hand side: $2x - y^{2}=\frac{dy}{dx}(4 + 2xy)$. Then $\frac{dy}{dx}=\frac{2x - y^{2}}{2xy + 4}$.

Step3: Find $y$ when $x = 5$

Substitute $x = 5$ into the original equation $x^{2}-xy^{2}=4y$, we have $25-5y^{2}=4y$. Rearrange it to the quadratic form $5y^{2}+4y - 25 = 0$. Using the quadratic formula $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 5$, $b = 4$, and $c=-25$, we get $y=\frac{-4\pm\sqrt{16+500}}{10}=\frac{-4\pm\sqrt{516}}{10}=\frac{-4\pm2\sqrt{129}}{10}=\frac{-2\pm\sqrt{129}}{5}$.

Step4: Find the slope of the tangent lines

When $y = y_1=\frac{-2+\sqrt{129}}{5}$ and $x = 5$, $\frac{dy}{dx}\big|{x = 5,y=\frac{-2+\sqrt{129}}{5}}=\frac{2\times5-(\frac{-2+\sqrt{129}}{5})^{2}}{2\times5\times\frac{-2+\sqrt{129}}{5}+4}$. When $y = y_2=\frac{-2-\sqrt{129}}{5}$ and $x = 5$, $\frac{dy}{dx}\big|{x = 5,y=\frac{-2-\sqrt{129}}{5}}=\frac{2\times5-(\frac{-2-\sqrt{129}}{5})^{2}}{2\times5\times\frac{-2-\sqrt{129}}{5}+4}$. The equation of a tangent line is $y - y_0=m(x - x_0)$. Here $x_0 = 5$. For $y_1=\frac{-2+\sqrt{129}}{5}$, the equation of the tangent line is $y-\frac{-2+\sqrt{129}}{5}=m_1(x - 5)$ where $m_1=\frac{10-(\frac{-2+\sqrt{129}}{5})^{2}}{2(-2+\sqrt{129})+4}$. For $y_2=\frac{-2-\sqrt{129}}{5}$, the equation of the tangent line is $y-\frac{-2-\sqrt{129}}{5}=m_2(x - 5)$ where $m_2=\frac{10-(\frac{-2-\sqrt{129}}{5})^{2}}{2(-2-\sqrt{129})+4}$. Simplifying the expression for $\frac{dy}{dx}$ from Step2:

Answer:

a) $\frac{dy}{dx}=\frac{2x - y^{2}}{2xy + 4}$ b) First, when $x = 5$, from $5y^{2}+4y - 25 = 0$, $y=\frac{-2\pm\sqrt{129}}{5}$. The slope $\frac{dy}{dx}=\frac{2x - y^{2}}{2xy + 4}$. For $y_1=\frac{-2+\sqrt{129}}{5}$, the tangent - line equation is $y-\frac{-2+\sqrt{129}}{5}=\frac{10 - (\frac{-2+\sqrt{129}}{5})^{2}}{10\times\frac{-2+\sqrt{129}}{5}+4}(x - 5)$. For $y_2=\frac{-2-\sqrt{129}}{5}$, the tangent - line equation is $y-\frac{-2-\sqrt{129}}{5}=\frac{10 - (\frac{-2-\sqrt{129}}{5})^{2}}{10\times\frac{-2-\sqrt{129}}{5}+4}(x - 5)$.