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Partial derivatives

Problem 10.127 · hard

For \( \displaystyle f(x, y) = 2 \operatorname{atan}{\left(\frac{y}{x} \right)} \), find \( \displaystyle f_x \), \( \displaystyle f_y \) and \( \displaystyle f_{xy} \).
  1. For f_x hold y constant; for f_y hold x constant.
  2. \[ \frac{\partial}{\partial x} 2 \operatorname{atan}{\left(\frac{y}{x} \right)} = - \frac{2 y}{x^{2} + y^{2}} \]
    f_x.✓ Proved
  3. \[ \frac{\partial}{\partial y} 2 \operatorname{atan}{\left(\frac{y}{x} \right)} = \frac{2 x}{x^{2} + y^{2}} \]
    f_y.✓ Proved
  4. \[ \frac{\partial}{\partial y} \left(- \frac{2 y}{x^{2} + y^{2}}\right) = \frac{- 2 x^{2} + 2 y^{2}}{\left(x^{2} + y^{2}\right)^{2}} \]
    f_xy: differentiate f_x with respect to y.✓ Proved
  5. \[ \frac{\partial}{\partial x} \frac{2 x}{x^{2} + y^{2}} = \frac{- 2 x^{2} + 2 y^{2}}{\left(x^{2} + y^{2}\right)^{2}} \]
    Clairaut: f_yx is the same.✓ Proved
Answer \( f_x = - \frac{2 y}{x^{2} + y^{2}},\quad f_y = \frac{2 x}{x^{2} + y^{2}},\quad f_{xy} = \frac{2 \left(- x^{2} + y^{2}\right)}{\left(x^{2} + y^{2}\right)^{2}} \)

Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.

The full receipt
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0difference quotients of f at (0.6, 0.4) agree with f_x and f_y

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/partial_derivatives, checked 2026-09-26 with SymPy 1.14.0.