Solución

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

f(x+h,y)=(x+h)2−3(x+h)y+2y2−4(x+h)+5y−12=x2+2xh+h2−3xy−3hy+2y2−4x−4h+5y−12\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

∂f∂x=lim⁡h→0f(x+h,y)−f(x,y)h=lim⁡h→0(x2+2xh+h2−3xy−3hy+2y2−4x−4h+5y−12)−(x2−3xy+2y2−4x+5y−12)h=lim⁡h→0x2+2xh+h2−3xy−3hy+2y2−4x−4h+5y−12−x2+3xy−2y2+4x−5y+12h=lim⁡h→02xh+h2−3hy−4hh=lim⁡h→0h(2x+h−3y−4)h=lim⁡h→0(2x+h−3y−4)=2x−3y−4\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 ∂f∂y\frac{\partial f}{\partial y}, primero calcula f(x,y+h)f (x, y + h)

f(x,y+h)=x2−3x(y+h)+2(y+h)2−4x+5(y+h)−12=x2−3xy−3xh+2y2+4yh+2h2−4x+5y+5h−12\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:

∂f∂y=lim⁡k→0f(x,y+k)−f(x,y)k=lim⁡k→0(x2−3xy−3xk+2y2+4yk+2k2−4x+5y+5k−12)−(x2−3xy+2y2−4x+5y−12)k=lim⁡k→0x2−3xy−3xk+2y2+4yk+2k2−4x+5y+5k−12−x2+3xy−2y2+4x−5y+12k=lim⁡k→0−3xk+4yk+2k2+5kk=lim⁡k→0k(−3x+4y+2k+5)k=lim⁡k→0(−3x+4y+2k+5)=−3x+4y+5\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}