# Solución

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

Apartado a

$$\begin{aligned} \color{blue} \frac{\partial f}{\partial x} &= \frac{\partial}{\partial x}\bigg[\frac{x^2y−4xz+y^2}{x-3yz}\bigg]\\ &= \frac{\frac{\partial}{\partial x}(x^2y−4xz+y^2)(x−3yz)−(x^2y−4xz+y^2)\frac{\partial}{\partial x}(x−3yz)}{(x−3yz)^2}\\ &= \frac{(2xy−4z)(x−3yz)−(x^2y−4xz+y^2)(1)}{(x−3yz)^2}\\ &= \frac{2x^2y−6xy^2z−4xz+12yz^2−x^2y+4xz-y^2}{(x−3yz)^2}\\ &= \frac{x^2y−6xy^2z−4xz+12yz^2+4xz−y^2}{(x−3yz)^2} \end{aligned}$$ $$\begin{aligned} \color{red} \frac{\partial f}{\partial y} &= \frac{\partial}{\partial y}\bigg[\frac{x^2y−4xz+y^2}{x-3yz}\bigg]\\ &= \frac{\frac{\partial}{\partial y}(x^2y−4xz+y^2)(x−3yz)−(x^2y−4xz+y^2)\frac{\partial}{\partial y}(x−3yz)}{(x−3yz)^2}\\ &= \frac{(x^2+2y)(x−3yz)−(x^2y−4xz+y^2)(−3z)}{(x−3yz)^2}\\ &= \frac{x^3−3x^2yz+2xy−6y^2z+3x^2yz−12xz^2+3y^2z}{(x−3yz)^2}\\ &= \frac{x^3+2xy−3y^2z−12xz^2}{(x−3yz)^2} \end{aligned}$$ $$\begin{aligned} \color{purple} \frac{\partial f}{\partial z} &= \frac{\partial}{\partial z} \bigg[\frac{x^2y−4xz+y^2}{x-3yz}\bigg]\\ &= \frac{\frac{\partial}{\partial z}(x^2y−4xz+y^2)(x−3yz)−(x^2y−4xz+y^2)\frac{\partial}{\partial z}(x−3yz)}{(x−3yz)^2}\\ &= \frac{(-4x)(x−3yz)−(x^2y−4xz+y^2)(−3y)}{(x−3yz)^2}\\ &= \frac{-4x^2+12xyz+3x^2y^2−12xyz+3y^3}{(x−3yz)^2}\\ &= \frac{-4x^2+3x^2y^2+3y^3}{(x−3yz)^2} \end{aligned}$$

Apartado b

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