Solución

En cada caso, trata todas las variables como constantes, excepto aquella cuya derivada parcial estás calculando.

Apartado a

∂f∂x=∂∂x[x2y−4xz+y2x−3yz]=∂∂x(x2y−4xz+y2)(x−3yz)−(x2y−4xz+y2)∂∂x(x−3yz)(x−3yz)2=(2xy−4z)(x−3yz)−(x2y−4xz+y2)(1)(x−3yz)2=2x2y−6xy2z−4xz+12yz2−x2y+4xz−y2(x−3yz)2=x2y−6xy2z−4xz+12yz2+4xz−y2(x−3yz)2\begin{aligned} \color{blue} \frac{\partial f}{\partial x} &= \frac{\partial}{\partial x}\bigg[\frac{x^2y−4xz+y^2}{x-3yz}\bigg]\\ &= \frac{\frac{\partial}{\partial x}(x^2y−4xz+y^2)(x−3yz)−(x^2y−4xz+y^2)\frac{\partial}{\partial x}(x−3yz)}{(x−3yz)^2}\\ &= \frac{(2xy−4z)(x−3yz)−(x^2y−4xz+y^2)(1)}{(x−3yz)^2}\\ &= \frac{2x^2y−6xy^2z−4xz+12yz^2−x^2y+4xz-y^2}{(x−3yz)^2}\\ &= \frac{x^2y−6xy^2z−4xz+12yz^2+4xz−y^2}{(x−3yz)^2} \end{aligned} ∂f∂y=∂∂y[x2y−4xz+y2x−3yz]=∂∂y(x2y−4xz+y2)(x−3yz)−(x2y−4xz+y2)∂∂y(x−3yz)(x−3yz)2=(x2+2y)(x−3yz)−(x2y−4xz+y2)(−3z)(x−3yz)2=x3−3x2yz+2xy−6y2z+3x2yz−12xz2+3y2z(x−3yz)2=x3+2xy−3y2z−12xz2(x−3yz)2\begin{aligned} \color{red} \frac{\partial f}{\partial y} &= \frac{\partial}{\partial y}\bigg[\frac{x^2y−4xz+y^2}{x-3yz}\bigg]\\ &= \frac{\frac{\partial}{\partial y}(x^2y−4xz+y^2)(x−3yz)−(x^2y−4xz+y^2)\frac{\partial}{\partial y}(x−3yz)}{(x−3yz)^2}\\ &= \frac{(x^2+2y)(x−3yz)−(x^2y−4xz+y^2)(−3z)}{(x−3yz)^2}\\ &= \frac{x^3−3x^2yz+2xy−6y^2z+3x^2yz−12xz^2+3y^2z}{(x−3yz)^2}\\ &= \frac{x^3+2xy−3y^2z−12xz^2}{(x−3yz)^2} \end{aligned} ∂f∂z=∂∂z[x2y−4xz+y2x−3yz]=∂∂z(x2y−4xz+y2)(x−3yz)−(x2y−4xz+y2)∂∂z(x−3yz)(x−3yz)2=(−4x)(x−3yz)−(x2y−4xz+y2)(−3y)(x−3yz)2=−4x2+12xyz+3x2y2−12xyz+3y3(x−3yz)2=−4x2+3x2y2+3y3(x−3yz)2\begin{aligned} \color{purple} \frac{\partial f}{\partial z} &= \frac{\partial}{\partial z} \bigg[\frac{x^2y−4xz+y^2}{x-3yz}\bigg]\\ &= \frac{\frac{\partial}{\partial z}(x^2y−4xz+y^2)(x−3yz)−(x^2y−4xz+y^2)\frac{\partial}{\partial z}(x−3yz)}{(x−3yz)^2}\\ &= \frac{(-4x)(x−3yz)−(x^2y−4xz+y^2)(−3y)}{(x−3yz)^2}\\ &= \frac{-4x^2+12xyz+3x^2y^2−12xyz+3y^3}{(x−3yz)^2}\\ &= \frac{-4x^2+3x^2y^2+3y^3}{(x−3yz)^2} \end{aligned}

Apartado b

∂f∂x=∂∂x[sen(x2y−z)+cos(x2−yz)]=cos(x2y−z)∂∂x(x2y−z)−sen(x2−yz)∂∂x(x2−yz)=2xycos(x2y−z)−2xsen(x2−yz)\begin{aligned} \color{blue} \frac{\partial f}{\partial x} &= \frac{\partial}{\partial x}[sen(x^2y−z)+cos(x^2−yz)]\\ &= cos(x^2y−z)\frac{\partial}{\partial x}(x^2y−z)−sen(x^2−yz)\frac{\partial}{\partial x}(x^2−yz)\\ &= 2xycos(x^2y−z)−2xsen(x^2−yz) \end{aligned} ∂f∂y=∂∂y[sen(x2y−z)+cos(x2−yz)]=(cos(x2y−z)∂∂y(x2y−z)−sen(x2−yz)∂∂y(x2−yz)=x2cos(x2y−z)+zsen(x2−yz)\begin{aligned} \color{red} \frac{\partial f}{\partial y} &= \frac{\partial}{\partial y}[sen(x^2y−z)+cos(x^2−yz)]\\ &= (cos(x^2y−z)\frac{\partial}{\partial y}(x^2 y−z)−sen(x^2−yz)\frac{\partial}{\partial y}(x^2−yz)\\ &= x^2cos(x^2y−z)+zsen(x^2−yz) \end{aligned} ∂f∂z=∂∂z[sen(x2y−z)+cos(x2−yz)]=cos(x2y−z)∂∂z(x2y−z)−sen(x2−yz)∂∂z(x2−yz)=−cos(x2y−z)+ysen(x2−yz)\begin{aligned} \color{brown} \frac{\partial f}{\partial z} &= \frac{\partial}{\partial z}[sen(x^2y−z)+cos(x^2−yz)]\\ &= cos(x^2y−z)\frac{\partial}{\partial z}(x^2y−z)−sen(x^2−yz)\frac{\partial}{\partial z}(x^2−yz)\\ &= −cos(x^2y−z)+ysen(x^2−yz) \end{aligned}