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

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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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