1. **Problem 1:** Given the function $$f(x,y,z) = x^2 y - 10 y^2 z^3 + 43 x - 7 \tan(5 y),$$ find the partial derivatives $$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}.$$ Step 1: Compute $$\frac{\partial f}{\partial x}$$ treating $$y,z$$ as constants. $$\frac{\partial}{\partial x}(x^2 y) = 2 x y$$ $$\frac{\partial}{\partial x}(-10 y^2 z^3) = 0$$ $$\frac{\partial}{\partial x}(43 x) = 43$$ $$\frac{\partial}{\partial x}(-7 \tan(5 y)) = 0$$ So, $$\frac{\partial f}{\partial x} = 2 x y + 43.$$ Step 2: Compute $$\frac{\partial f}{\partial y}$$ treating $$x,z$$ as constants. $$\frac{\partial}{\partial y}(x^2 y) = x^2$$ $$\frac{\partial}{\partial y}(-10 y^2 z^3) = -10 \cdot 2 y z^3 = -20 y z^3$$ $$\frac{\partial}{\partial y}(43 x) = 0$$ $$\frac{\partial}{\partial y}(-7 \tan(5 y)) = -7 \cdot \sec^2(5 y) \cdot 5 = -35 \sec^2(5 y)$$ So, $$\frac{\partial f}{\partial y} = x^2 - 20 y z^3 - 35 \sec^2(5 y).$$ Step 3: Compute $$\frac{\partial f}{\partial z}$$ treating $$x,y$$ as constants. $$\frac{\partial}{\partial z}(x^2 y) = 0$$ $$\frac{\partial}{\partial z}(-10 y^2 z^3) = -10 y^2 \cdot 3 z^2 = -30 y^2 z^2$$ $$\frac{\partial}{\partial z}(43 x) = 0$$ $$\frac{\partial}{\partial z}(-7 \tan(5 y)) = 0$$ So, $$\frac{\partial f}{\partial z} = -30 y^2 z^2.$$ --- 2. **Problem 2:** Check if $$f(x,y) = \cos\left(x - \frac{y}{3}\right)$$ satisfies Laplace's equation $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0.$$ Step 1: Compute $$\frac{\partial f}{\partial x} = -\sin\left(x - \frac{y}{3}\right)$$ Step 2: Compute $$\frac{\partial^2 f}{\partial x^2} = -\cos\left(x - \frac{y}{3}\right)$$ Step 3: Compute $$\frac{\partial f}{\partial y} = -\sin\left(x - \frac{y}{3}\right) \cdot \left(-\frac{1}{3}\right) = \frac{1}{3} \sin\left(x - \frac{y}{3}\right)$$ Step 4: Compute $$\frac{\partial^2 f}{\partial y^2} = \frac{1}{3} \cos\left(x - \frac{y}{3}\right) \cdot \left(-\frac{1}{3}\right) = -\frac{1}{9} \cos\left(x - \frac{y}{3}\right)$$ Step 5: Sum the second derivatives: $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = -\cos\left(x - \frac{y}{3}\right) - \frac{1}{9} \cos\left(x - \frac{y}{3}\right) = -\frac{10}{9} \cos\left(x - \frac{y}{3}\right) \neq 0.$$ Conclusion: The function does not satisfy Laplace's equation. --- 3. **Problem 3:** Given $$w = x^2 y + y^2,$$ with $$x(t) = e^{5 t}$$ and $$y(t) = \sin t,$$ find $$\frac{d w}{d t}$$ using the chain rule and simplify. Step 1: Compute partial derivatives: $$\frac{\partial w}{\partial x} = 2 x y$$ $$\frac{\partial w}{\partial y} = x^2 + 2 y$$ Step 2: Compute derivatives of $$x(t)$$ and $$y(t)$$: $$\frac{d x}{d t} = 5 e^{5 t}$$ $$\frac{d y}{d t} = \cos t$$ Step 3: Apply chain rule: $$\frac{d w}{d t} = \frac{\partial w}{\partial x} \frac{d x}{d t} + \frac{\partial w}{\partial y} \frac{d y}{d t} = 2 x y (5 e^{5 t}) + (x^2 + 2 y) \cos t.$$ Step 4: Substitute $$x = e^{5 t}$$ and $$y = \sin t$$: $$\frac{d w}{d t} = 2 e^{5 t} \sin t (5 e^{5 t}) + (e^{10 t} + 2 \sin t) \cos t = 10 e^{10 t} \sin t + e^{10 t} \cos t + 2 \sin t \cos t.$$ Final simplified derivative: $$\frac{d w}{d t} = 10 e^{10 t} \sin t + e^{10 t} \cos t + 2 \sin t \cos t.$$