Partial Derivatives — Question 2

Question 2 – Solution

Strategy. Apply the one-variable quotient rule separately in each variable and preserve the original domain.

Step 1: Domain The function and its partial derivatives are defined where $$\boxed{2x-y\ne 0}.$$

Step 2: $x$-partial Holding $y$ fixed, $$\begin{align*} g_x&=\frac{(2x)(2x-y)-2(x^2+3y)}{(2x-y)^2}\\ &=\boxed{\frac{2x^2-2xy-6y}{(2x-y)^2}}. \end{align*}$$

Step 3: $y$-partial Holding $x$ fixed, $$\begin{align*} g_y&=\frac{3(2x-y)-(x^2+3y)(-1)}{(2x-y)^2}\\ &=\boxed{\frac{x^2+6x}{(2x-y)^2}}. \end{align*}$$

Verification. Both results have the squared quotient-rule denominator and remain restricted to $2x-y\ne 0$; expanding each unsimplified numerator reproduces the displayed form.