if $x^{2}+xy = 10$, then when $x = 2$, $\frac{dy}{dx}=$\na $-\frac{7}{2}$\nb -2\nc $\frac{2}{7}$\nd…

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})