Question

calculate $\frac{dy}{dx}$ for $3x + xy = y$ and find the instantaneous rate of change at the point $(\frac{1}{4},1)$.

Explanation:

Step1: Differentiate both sides

Differentiate $3x+xy = y$ with respect to $x$. The derivative of $3x$ is $3$, for $xy$ use product - rule $(uv)^\prime=u^\prime v + uv^\prime$ where $u = x$ and $v = y$, so $(xy)^\prime=y+xy^\prime$, and the derivative of $y$ is $y^\prime$. We get $3 + y+xy^\prime=y^\prime$.

Step2: Solve for $y^\prime$

Rearrange the equation $3 + y+xy^\prime=y^\prime$ to isolate $y^\prime$. First, move all terms with $y^\prime$ to one side: $xy^\prime - y^\prime=-3 - y$. Then factor out $y^\prime$: $y^\prime(x - 1)=-(3 + y)$. So $y^\prime=\frac{3 + y}{1 - x}$.

Step3: Find the value at the given point

Substitute $x=\frac{1}{4}$ and $y = 1$ into $y^\prime$. $y^\prime=\frac{3+1}{1-\frac{1}{4}}=\frac{4}{\frac{3}{4}}=\frac{16}{3}$.

Answer:

$\frac{16}{3}$