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

Question

Accepted 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:
$-\frac{4}{5}$

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QUESTION IMAGE

Question

Explanation:

Step1: Differentiate both sides with respect to \(x\)

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

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

Answer: