Partial Derivatives of x^3 y^2 - 2xy^4

Two monomials, two exponents each

Every term here contains both variables, so neither partial loses a term. For $\frac{\partial f}{\partial x}$, the $y$ powers stay exactly as they are and only the $x$ powers move.

Notice what happened to the second term: $-2xy^4$ has $x$ to the first power, so differentiating it leaves $-2y^4$ with no $x$ at all.

For $\frac{\partial f}{\partial y}$, the $x$ powers freeze and the $y$ exponents come down.

The coefficient $-8$ comes from $-2 \cdot 4$, the original coefficient times the exponent on $y$. At $(1,1)$ the partials are $1$ and $-6$.

The mistake: multiplying the exponents together

A frequent wrong answer for the first term of $\frac{\partial f}{\partial x}$ is $3x^2 \cdot 2y$, formed by differentiating both exponents in the same step. Only one variable is active per partial derivative.

• In $\frac{\partial}{\partial x}\left(x^3y^2\right)$, the exponent $3$ comes down and $y^2$ is untouched, giving $3x^2y^2$.
• In $\frac{\partial}{\partial y}\left(x^3y^2\right)$, the exponent $2$ comes down and $x^3$ is untouched, giving $2x^3y$.
• Bringing both exponents down at once would be the mixed second partial, a different object entirely.

That mixed partial is worth naming, because it is where the two exponents legitimately meet: $f_{xy} = \frac{\partial}{\partial y}\left(3x^2y^2 - 2y^4\right) = 6x^2y - 8y^3$.

Cross-check with Clairaut's theorem

For a polynomial, the mixed partials must be equal no matter which order you differentiate in. That gives you a free check on both answers at once.

They match, so any error would have to be one that survives a second differentiation in both orders, which is rare. If they had disagreed, at least one of the first partials would be wrong and you would know to redo them before going further.