1. **Problema 1:** Dada la función $$w = e^x \sin y + y^y \sin x$$ con $$x = 3t$$ y $$y = 2t$$, encontrar $$\frac{dw}{dt}$$ cuando $$t = \frac{\sqrt{3}}{2}$$. 2. **Fórmula y regla usada:** Para funciones compuestas, la regla de la cadena dice que $$\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt}$$. 3. **Calcular derivadas parciales:** $$\frac{\partial w}{\partial x} = e^x \sin y + y^y \cos x$$ Para $$\frac{\partial w}{\partial y}$$, usar que $$\frac{d}{dy} y^y = y^y (\ln y + 1)$$: $$\frac{\partial w}{\partial y} = e^x \cos y + y^y (\ln y + 1) \sin x$$ 4. **Derivar $$x$$ y $$y$$ respecto a $$t$$:** $$\frac{dx}{dt} = 3$$ $$\frac{dy}{dt} = 2$$ 5. **Evaluar en $$t = \frac{\sqrt{3}}{2}$$:** $$x = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$$ $$y = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$$ 6. **Calcular $$\frac{\partial w}{\partial x}$$ y $$\frac{\partial w}{\partial y}$$ en esos valores:** $$\frac{\partial w}{\partial x} = e^{\frac{3\sqrt{3}}{2}} \sin(\sqrt{3}) + (\sqrt{3})^{\sqrt{3}} \cos\left(\frac{3\sqrt{3}}{2}\right)$$ $$\frac{\partial w}{\partial y} = e^{\frac{3\sqrt{3}}{2}} \cos(\sqrt{3}) + (\sqrt{3})^{\sqrt{3}} (\ln(\sqrt{3}) + 1) \sin\left(\frac{3\sqrt{3}}{2}\right)$$ 7. **Finalmente:** $$\frac{dw}{dt} = \frac{\partial w}{\partial x} \times 3 + \frac{\partial w}{\partial y} \times 2$$ --- **Problema 2:** Dada $$w = y^3 - 3x^2 y$$ con $$x = e^t$$ y $$y = e^t$$, encontrar $$\frac{dw}{dt}$$. 1. Usamos la regla de la cadena: $$\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt}$$ 2. Derivadas parciales: $$\frac{\partial w}{\partial x} = -6xy$$ $$\frac{\partial w}{\partial y} = 3y^2 - 3x^2$$ 3. Derivadas de $$x$$ y $$y$$: $$\frac{dx}{dt} = e^t$$ $$\frac{dy}{dt} = e^t$$ 4. Evaluar en $$x = e^t$$, $$y = e^t$$: $$\frac{dw}{dt} = (-6 e^t e^t)(e^t) + (3 (e^t)^2 - 3 (e^t)^2)(e^t) = -6 e^{3t} + 0 = -6 e^{3t}$$ --- **Problema 3:** Dada $$w = 3x^3 y + y^3$$ con $$x = s + t$$ y $$y = s - 2t$$, encontrar $$\frac{dw}{dt}$$ y $$\frac{dw}{ds}$$. 1. Regla de la cadena: $$\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt}$$ $$\frac{dw}{ds} = \frac{\partial w}{\partial x} \frac{dx}{ds} + \frac{\partial w}{\partial y} \frac{dy}{ds}$$ 2. Derivadas parciales: $$\frac{\partial w}{\partial x} = 9x^2 y$$ $$\frac{\partial w}{\partial y} = 3x^3 + 3y^2$$ 3. Derivadas de $$x$$ y $$y$$: $$\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = -2$$ $$\frac{dx}{ds} = 1, \quad \frac{dy}{ds} = 1$$ 4. Sustituir: $$\frac{dw}{dt} = 9x^2 y (1) + (3x^3 + 3y^2)(-2) = 9x^2 y - 6x^3 - 6y^2$$ $$\frac{dw}{ds} = 9x^2 y (1) + (3x^3 + 3y^2)(1) = 9x^2 y + 3x^3 + 3y^2$$ 5. Expresar en términos de $$s$$ y $$t$$: $$x = s + t, \quad y = s - 2t$$ Por lo tanto: $$\frac{dw}{dt} = 9 (s+t)^2 (s - 2t) - 6 (s+t)^3 - 6 (s - 2t)^2$$ $$\frac{dw}{ds} = 9 (s+t)^2 (s - 2t) + 3 (s+t)^3 + 3 (s - 2t)^2$$