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ln⁡(x)+ln⁡(y)−ln⁡(xy)\ln(x) + \ln(y) - \ln(xy)

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Full solution

Q. ln⁡(x)+ln⁡(y)−ln⁡(xy)\ln(x) + \ln(y) - \ln(xy)
  1. Apply Logarithm Properties: Apply the properties of logarithms to simplify the function. The properties of logarithms tell us that ln⁡(a)+ln⁡(b)=ln⁡(ab)\ln(a) + \ln(b) = \ln(ab) and ln⁡(a/b)=ln⁡(a)−ln⁡(b)\ln(a/b) = \ln(a) - \ln(b). Using these properties, we can simplify the function f(x,y)f(x, y) as follows: f(x,y)=ln⁡(x)+ln⁡(y)−ln⁡(xy)=ln⁡(x)+ln⁡(y)−ln⁡(x)−ln⁡(y)=0f(x, y) = \ln(x) + \ln(y) - \ln(xy) = \ln(x) + \ln(y) - \ln(x) - \ln(y) = 0.
  2. Derivative of Constant Function: Since the simplified function f(x,y)f(x, y) is a constant (00), its derivative with respect to xx or yy is also a constant. The derivative of a constant is 00. Therefore, the partial derivatives of f(x,y)f(x, y) with respect to xx and yy are: ∂f/∂x=0\partial f/\partial x = 0 and ∂f/∂y=0\partial f/\partial y = 0.
  3. Write Gradient of Function: Combine the partial derivatives to write the gradient of f(x,y)f(x, y). The gradient of f(x,y)f(x, y) is the vector of its partial derivatives, which in this case is: \(\newlineabla f(x, y) = (0, 0)\).

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