Let's solve the given question in **2 steps** with explanations: ### Given: - \( w = f(u) + g(v) \) - \( u = x + iy \) - \( v = x - iy \) - \( i = \sqrt{-1} \) We are to show: \[ w_{xx} + w_{yy} = 0 \] --- ## **Step 1: Find the first and second partial derivatives of \( w \) with respect to \( x \) and \( y \)** First, compute the derivatives of \( u \) and \( v \): \[ \frac{\partial u}{\partial x} = 1,\quad \frac{\partial u}{\partial y} = i \] \[ \frac{\partial v}{\partial x} = 1,\quad \frac{\partial v}{\partial y} = -i \] Now, the first derivatives: \[ \frac{\partial w}{\partial x} = f'(u)\frac{\partial u}{\partial x} + g'(v)\frac{\partial v}{\partial x} = f'(u) \cdot 1 + g'(v) \cdot 1 = f'(u) + g'(v) \] \[ \frac{\partial w}{\partial y} = f'(u)\frac{\partial u}{\partial y} + g'(v)\frac{\partial v}{\partial y} = f'(u) \cdot i + g'(v) \cdot (-i) = i f'(u) - i g'(v) \] **Explanation:** We use the chain rule for partial derivatives, applying to both \( f(u) \) and \( g(v) \). --- ## **Step 2: Compute \( w_{xx} \) and \( w_{yy} \), then add** Now, let's find the second derivatives: \[ w_{xx} = \frac{\partial}{\partial x} \left( f'(u) + g'(v) \right) \] \[ = f''(u) \frac{\partial u}{\partial x} + g''(v) \frac{\partial v}{\partial x} = f''(u) \cdot 1 + g''(v) \cdot 1 = f''(u) + g''(v) \] Similarly, \[ w_{yy} = \frac{\partial}{\partial y} \left( i f'(u) - i g'(v) \right) \] \[ = i f''(u) \frac{\partial u}{\partial y} - i g''(v) \frac{\partial v}{\partial y} \] \[ = i f''(u) \cdot i - i g''(v) \cdot (-i) = i^2 f''(u) + i^2 g''(v) = (-1) f''(u) + (-1) g''(v) = -f''(u) - g''(v) \] Now add: \[ w_{xx} + w_{yy} = (f''(u) + g''(v)) + (-f''(u) - g''(v)) = 0 \] **Explanation:** The second derivative terms cancel out exactly, so the sum is zero. --- ## **Final Answer:** \[ \boxed{w_{xx} + w_{yy} = 0} \] This shows that the function \( w \) satisfies the 2D Laplace equation.