سوال ۳۲ کنکور ارشد مهندسی کامپیوتر ۱۴۰۰

Let $u = \dfrac{y^{2}}{x}$, so $z = x^{n}e^{u}$ with $\dfrac{\partial u}{\partial y} = \dfrac{2y}{x}$ and $\dfrac{\partial u}{\partial x} = -\dfrac{y^{2}}{x^{2}}$.

Step 1 — the $y$-derivative. $\frac{\partial z}{\partial y} = x^{n}e^{y^{2}/x}\cdot\frac{2y}{x} = 2y\,x^{n-1}e^{y^{2}/x}.$ Hence $-y^{3}\frac{\partial z}{\partial y} = -2y^{4}\,x^{n-1}e^{y^{2}/x}.$

Step 2 — differentiate again in $y$ (product rule, treating $x$ as constant): $\frac{\partial}{\partial y}\left(-2y^{4}x^{n-1}e^{y^{2}/x}\right) = -2x^{n-1}e^{y^{2}/x}\left(4y^{3} + y^{4}\cdot\frac{2y}{x}\right) = -2x^{n-1}e^{y^{2}/x}\left(4y^{3} + \frac{2y^{5}}{x}\right).$

Step 3 — multiply by $\dfrac{1}{4y^{3}}$: $\frac{1}{4y^{3}}\cdot\left(-2x^{n-1}e^{y^{2}/x}\right)\left(4y^{3} + \frac{2y^{5}}{x}\right) = -x^{n-1}e^{y^{2}/x}\left(2 + \frac{y^{2}}{x}\right) = -2x^{n-1}e^{y^{2}/x} - y^{2}x^{n-2}e^{y^{2}/x}.$ This is the left-hand side.

Step 4 — the $x$-derivative (right-hand side): $\frac{\partial z}{\partial x} = n x^{n-1}e^{y^{2}/x} + x^{n}e^{y^{2}/x}\cdot\left(-\frac{y^{2}}{x^{2}}\right) = n x^{n-1}e^{y^{2}/x} - y^{2}x^{n-2}e^{y^{2}/x}.$

Step 5 — match. Setting LHS $=$ RHS, the term $-y^{2}x^{n-2}e^{y^{2}/x}$ cancels from both sides, leaving $-2x^{n-1}e^{y^{2}/x} = n\,x^{n-1}e^{y^{2}/x}\;\Longrightarrow\; n = -2.$

So the identity holds exactly when $n = -2$ (option 4).

Exam trick: Factor out the common $e^{y^{2}/x}$ early and compare the two sides term by term — the $y^{2}x^{n-2}$ pieces are built to cancel, so the answer is fixed by the single $x^{n-1}$ coefficient $n = -2$.