# Solución

Para calcular $\frac{\partial^2 f}{\partial x^2}$ y $\frac{\partial^2 f}{\partial y\partial x}$, primero calculamos $\frac{\partial f}{\partial x}$:

$$\frac{\partial f}{\partial x} = e^{−3y}+2cos(2x−5y)$$

Para calcular $\frac{\partial^2 f}{\partial x^2}$, diferencia $\frac{\partial f}{\partial x}$ con respecto a $x$:

$$\begin{aligned} \frac{\partial^2 f}{\partial x^2} &= \frac{\partial}{\partial x}\bigg[\frac{\partial f}{\partial x} \bigg]\\ &= \frac{\partial}{\partial x}\big[e^{−3y}+2cos(2x−5y) \big]\\ &= −4sen(2x−5y) \end{aligned}$$

Para calcular $\frac{\partial^2 f}{\partial y\partial x}$, diferencia $\frac{\partial f}{\partial x}$ con respecto a $y$:

$$\begin{aligned} \frac{\partial^2 f}{\partial y\partial x} &= \frac{\partial}{\partial y}\bigg[\frac{\partial f}{\partial x}\bigg]\\ &= \frac{\partial}{\partial y}\big[e^{−3y}+2cos(2x−5y) \big]\\ &= −3e^{−3y}+10sen(2x−5y) \end{aligned}$$

Para calcular $\frac{\partial^2 f}{\partial x\partial y}$ y $\frac{\partial^2 f}{\partial y^2}$, primero calculamos $\frac{\partial f}{\partial y}$:

$$\frac{\partial f}{\partial y} = −3xe^{−3y}−5cos(2x−5y)$$

Para calcular $\frac{\partial^2 f}{\partial x\partial y}$, diferencia $\frac{\partial f}{\partial y}$ con respecto a $x$:

$$\begin{aligned} \frac{\partial^2 f}{\partial x\partial y} &= \frac{\partial}{\partial x}\bigg[\frac{\partial f}{\partial y} \bigg]\\ &= \frac{\partial}{\partial x}\big[−3xe^{−3y}−5cos(2x−5y) \big]\\ &= −3e^{−3y}-10sen(2x−5y) \end{aligned}$$

Para calcular $\frac{\partial^2 f}{\partial y^2}$, diferencia $\frac{\partial f}{\partial y}$ con respecto a $y$:

$$\begin{aligned} \frac{\partial^2 f}{\partial y^2} &= \frac{\partial}{\partial y}\bigg[\frac{\partial f}{\partial y} \bigg]\\ &= \frac{\partial}{\partial y}\big[−3xe^{−3y}−5cos(2x−5y)\big]\\ &= 9xe^{−3y}+25sen(2x−5y) \end{aligned}$$