Solución

Para calcular ∂2f∂x2\frac{\partial^2 f}{\partial x^2} y ∂2f∂y∂x\frac{\partial^2 f}{\partial y\partial x}, primero calculamos ∂f∂x\frac{\partial f}{\partial x}:

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

Para calcular ∂2f∂x2\frac{\partial^2 f}{\partial x^2}, diferencia ∂f∂x\frac{\partial f}{\partial x} con respecto a xx:

∂2f∂x2=∂∂x[∂f∂x]=∂∂x[e−3y+2cos(2x−5y)]=−4sen(2x−5y)\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 ∂2f∂y∂x\frac{\partial^2 f}{\partial y\partial x}, diferencia ∂f∂x\frac{\partial f}{\partial x} con respecto a yy:

∂2f∂y∂x=∂∂y[∂f∂x]=∂∂y[e−3y+2cos(2x−5y)]=−3e−3y+10sen(2x−5y)\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 ∂2f∂x∂y\frac{\partial^2 f}{\partial x\partial y} y ∂2f∂y2\frac{\partial^2 f}{\partial y^2}, primero calculamos ∂f∂y\frac{\partial f}{\partial y}:

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

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

∂2f∂x∂y=∂∂x[∂f∂y]=∂∂x[−3xe−3y−5cos(2x−5y)]=−3e−3y−10sen(2x−5y)\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 ∂2f∂y2\frac{\partial^2 f}{\partial y^2}, diferencia ∂f∂y\frac{\partial f}{\partial y} con respecto a yy:

∂2f∂y2=∂∂y[∂f∂y]=∂∂y[−3xe−3y−5cos(2x−5y)]=9xe−3y+25sen(2x−5y)\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}