# Solución

Primero, calcula $f(x+h,y)$.

$$\begin{aligned} f(x+h,y) &= (x+h)^2−3(x+h)y+2y^2−4(x+h)+5y−12\\ &= x^2+2xh+h^2−3xy−3hy+2y^2−4x−4h+5y−12 \end{aligned}$$

Ahora, sustituye loa anterior en la ecuación 4.12 y simplifica

$$\begin{aligned} \frac{\partial f}{\partial x} &= \lim\limits_{h \to 0}\frac{f(x+h,y)−f(x,y)}{h}\\ &= \lim\limits_{h \to 0}\frac{(x^2+2xh+h^2−3xy−3hy+2y^2−4x−4h+5y−12)−(x^2−3xy+2y^2−4x+5y−12)}{h}\\ &= \lim\limits_{h \to 0}\frac{x^2+2xh+h^2−3xy−3hy+2y^2−4x−4h+5y−12−x^2+3xy−2y^2+4x−5y+12}{h}\\ &= \lim\limits_{h \to 0}\frac{2xh+h^2−3hy−4h}{h}\\ &= \lim\limits_{h \to 0}\frac{h(2x+h−3y−4)}{h}\\ &= \lim\limits_{h \to 0}(2x+h−3y−4)\\ &= 2x-3y-4 \end{aligned}$$

Para calcular $\frac{\partial f}{\partial y}$, primero calcula $f (x, y + h)$

$$\begin{aligned} f(x,y+h) &= x^2−3x(y+h)+2(y+h)^2−4x+5(y+h)−12\\ &= x^2−3xy−3xh+2y^2+4yh+2h^2−4x+5y+5h−12 \end{aligned}$$

Luego, sustituye esto en la ecuación 4.13 y simplifica:

$$\begin{aligned} \frac{\partial f}{\partial y} &= \lim\limits_{k \to 0}\frac{f(x,y+k)−f(x,y)}{k}\\ &= \lim\limits_{k \to 0}\frac{(x^2−3xy−3xk+2y^2+4yk+2k^2−4x+5y+5k−12)−(x^2−3xy+2y^2−4x+5y−12)}{k}\\ &= \lim\limits_{k \to 0}\frac{x^2−3xy−3xk+2y^2+4yk+2k^2−4x+5y+5k−12−x^2+3xy−2y^2+4x−5y+12}{k}\\ &= \lim\limits_{k \to 0}\frac{−3xk+4yk+2k^2+5k}{k}\\ &= \lim\limits_{k \to 0}\frac{k(−3x+4y+2k+5)}{k}\\ &= \lim\limits_{k \to 0}(−3x+4y+2k+5)\\ &= −3x+4y+5 \end{aligned}$$