Finding Partials of f(x,y) = x³y² - 2xy + y³

Calculus 3 · Axiom Academy

EXAMPLE Finding Partials of f(x,y) = x^3y^2 - 2xy + y^3 Compute and by treating the other variable as a constant. Find both first-order partial derivatives of f(x,y) = x^3y^2 - 2xy + y^3 . To take , treat y as a constant and differentiate term by term; then reverse the roles for . Nice work — you found both partials by freezing one variable at a time. Here is what to carry forward: Freeze the other variable: for , treat y (and every power of y ) as a constant; for , treat x as a constant. Constant terms vanish: in the pure- y term y^3 has no x , so its derivative is 0 ; in that same term becomes 3y^2 . Second partials & Clairaut: differentiating again gives f_ xx = 6xy^2 , f_ yy = 2x^3 + 6y , and the mixed partials agree, f_ xy = f_ yx = 6x^2y - 2 . Partial derivatives are the foundation for gradients, tangent planes, and multivariable optimization — the same freeze-one-variable move works for functions of any number of variables.

This is the written version of the interactive lesson above. See the full Calculus 3 course.