find the value of $\\frac{dy}{dx}$ at the point $(2,1)$. $x^{3}y + y^{2}-x^{2}=5$ choose 1 answer:

Explanation:

Step1: Differentiate both sides with respect to (x)

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = x^{3}), (v = y) and ((y^{2})^\prime=2y\frac{dy}{dx}), ((x^{2})^\prime = 2x) and ((5)^\prime=0). Differentiating (x^{3}y + y^{2}-x^{2}=5) gives: [ \begin{align*} 3x^{2}y+x^{3}\frac{dy}{dx}+2y\frac{dy}{dx}-2x&=0 \end{align*} ]

Step2: Solve for (\frac{dy}{dx})

Group the terms with (\frac{dy}{dx}): [x^{3}\frac{dy}{dx}+2y\frac{dy}{dx}=2x - 3x^{2}y] Factor out (\frac{dy}{dx}): [\frac{dy}{dx}(x^{3}+2y)=2x - 3x^{2}y] Then (\frac{dy}{dx}=\frac{2x - 3x^{2}y}{x^{3}+2y})

Step3: Substitute (x = 2) and (y = 1)

[ \begin{align*} \frac{dy}{dx}\big|_{(x = 2,y = 1)}&=\frac{2\times2-3\times2^{2}\times1}{2^{3}+2\times1}\ &=\frac{4 - 12}{8 + 2}\ &=\frac{-8}{10}\ &=-\frac{4}{5} \end{align*} ]

Answer:

(-\frac{4}{5})