start leveling up and building your weekly streak! 0 week streak level 1 ① (x^{3}y + y^{2}-x^{2}=5) find the…

Explanation:

Step1: Differentiate both sides

Differentiate $x^{3}y + y^{2}-x^{2}=5$ with respect to $x$ using product - rule and chain - rule. The product - rule states that $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{3}$ and $v = y$. The derivative of $x^{3}y$ is $3x^{2}y+x^{3}\frac{dy}{dx}$, the derivative of $y^{2}$ is $2y\frac{dy}{dx}$, and the derivative of $-x^{2}$ is $-2x$, and the derivative of the constant 5 is 0. So we have $3x^{2}y+x^{3}\frac{dy}{dx}+2y\frac{dy}{dx}-2x = 0$.

Step2: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side: $x^{3}\frac{dy}{dx}+2y\frac{dy}{dx}=2x - 3x^{2}y$. 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 the point $(2,1)$

Substitute $x = 2$ and $y = 1$ into the derivative formula: $\frac{dy}{dx}=\frac{2\times2-3\times2^{2}\times1}{2^{3}+2\times1}=\frac{4 - 12}{8 + 2}=\frac{-8}{10}=-\frac{4}{5}$.

Answer:

A. $-\frac{4}{5}$