Partial Derivatives (Chapter 12.3)

Question 12)
\(f(x, y) = \frac{4}{xy}\). Find \(f_{x}\), \(f_{y}\), \(f_{xx}\), \(f_{yy}\), \(f_{xy}\), \(f_{yx}.\)
Solution:

\(f(x, y) = \frac{4}{xy}\)

  1. \(f_{x}(x, y) = \frac{\partial}{\partial x}(\frac{4}{xy}) = \frac{4}{y}.\frac{d}{dx}.(\frac{1}{x}) = \frac{4}{y}.\frac{-1}{x^2} = -\frac{4}{x^2y}\)

  2. \(f_{y}(x, y) = \frac{\partial}{\partial y}(\frac{4}{xy}) = \frac{4}{x}.\frac{d}{dy}.(\frac{1}{y}) = \frac{4}{x}.\frac{-1}{y^2} = -\frac{4}{xy^2}\)

  3. \(f_{xx}(x, y) = \frac{\partial}{\partial x}(f_{x}) = \frac{\partial}{\partial x}(-\frac{4}{x^2y}) = \frac{-4}{y}.\frac{-2}{x^3} = \frac{8}{x^3y}\)

  4. \(f_{yy}(x, y) = \frac{\partial}{\partial y}(f_{y}) = \frac{\partial}{\partial y}(-\frac{4}{xy^2}) = \frac{-4}{x}.\frac{-2}{y^3} = \frac{8}{xy^3}\)

  5. \(f_{xy}(x, y) = \frac{\partial}{\partial y}(f_{x}) = \frac{\partial}{\partial y} (-\frac{4}{x^2y}) = \frac{-4}{x^2}. \frac{-1}{y^2} = \frac{4}{x^2y^2}\)

  6. \(f_{yx}(x, y) = \frac{\partial}{\partial x}(f_{y}) = \frac{\partial}{\partial x} (-\frac{4}{xy^2}) = \frac{-4}{y^2}. \frac{-1}{x^2} = \frac{4}{x^2y^2}\)

Separator - 01
Question 23)
\(f(x, y) = 3x^2+1\). Find \(f_{x}\), \(f_{y}\), \(f_{xx}\), \(f_{yy}\), \(f_{xy}\), \(f_{yx}.\)
Solution:

\(f(x, y) = 3x^2+1\)

  1. \(f_{x}(x, y) = \frac{\partial}{\partial x}(3x^2+1) = 6x\)

  2. \(f_{y}(x, y) = \frac{\partial}{\partial y}(3x^2+1) = (3x^2.0 + 0) = 0\)

  3. \(f_{xx}(x, y) = \frac{\partial}{\partial x}(f_{x}) = \frac{\partial}{\partial x}(6x) = 6\)

  4. \(f_{yy}(x, y) = \frac{\partial}{\partial y}(f_{y}) = \frac{\partial}{\partial y}(0) = 0\)

  5. \(f_{xy}(x, y) = \frac{\partial}{\partial y}(f_{x}) = \frac{\partial}{\partial y} (6x) = 6x.(0) = 0\)

  6. \(f_{yx}(x, y) = \frac{\partial}{\partial x}(f_{y}) = \frac{\partial}{\partial x} (0) = 0\)



What were the most valuable elements you took away from this course?

This course was a good review of linear algebra, probability, statistic and calculus which is very important in the field of Data Science.