Question

if $x^{2}+xy = 10$, then when $x = 2$, $\frac{dy}{dx}=$
a $-\frac{7}{2}$
b -2
c $\frac{2}{7}$
d $\frac{3}{2}$

Explanation:

Step1: Find the value of \(y\) when \(x = 2\)

Substitute \(x = 2\) into \(x^{2}+xy = 10\), we get \(2^{2}+2y=10\), which simplifies to \(4 + 2y=10\). Then \(2y=10 - 4=6\), so \(y = 3\).

Step2: Differentiate the equation \(x^{2}+xy = 10\) with respect to \(x\)

Using the sum rule and product rule. The derivative of \(x^{2}\) is \(2x\). For \(xy\), by the product rule \((uv)^\prime=u^\prime v+uv^\prime\) (where \(u = x\), \(v = y\)), its derivative is \(y+x\frac{dy}{dx}\). The derivative of the constant \(10\) is \(0\). So \(2x+y+x\frac{dy}{dx}=0\).

Step3: Solve for \(\frac{dy}{dx}\)

Rearrange \(2x+y+x\frac{dy}{dx}=0\) to get \(x\frac{dy}{dx}=-2x - y\), then \(\frac{dy}{dx}=\frac{-2x - y}{x}\).

Step4: Substitute \(x = 2\) and \(y = 3\) into \(\frac{dy}{dx}\)

\(\frac{dy}{dx}=\frac{-2\times2-3}{2}=\frac{-4 - 3}{2}=-\frac{7}{2}\)

Answer:

A. \(-\frac{7}{2}\)