Multivariable calculus

Partial Derivatives of (x^2 y + 1)^3

For f(x, y) = (x^2 y + 1)^3 the partial with respect to x is 6xy(x^2 y + 1)^2, and the partial with respect to y is 3x^2 (x^2 y + 1)^2. Both come from the chain rule: differentiate the cube to get 3(x^2 y + 1)^2, then multiply by the partial of the inside, which is 2xy for x and x^2 for y.

f(x,y)=(x2y+1)3f(x,y) = \left(x^2 y + 1\right)^3
∂f∂x=6xy(x2y+1)2\frac{\partial f}{\partial x} = 6xy\left(x^2 y + 1\right)^2
∂f∂y=3x2(x2y+1)2\frac{\partial f}{\partial y} = 3x^2\left(x^2 y + 1\right)^2

One outer derivative, two different insides

Write u=x2y+1u = x^2 y + 1, so f=u3f = u^3. The outer derivative ddu(u3)=3u2\frac{d}{du}\left(u^3\right) = 3u^2 is the same no matter which variable you differentiate with respect to. Only the inner factor changes.

∂f∂x=3u2∂u∂x,∂f∂y=3u2∂u∂y\frac{\partial f}{\partial x} = 3u^2 \frac{\partial u}{\partial x}, \qquad \frac{\partial f}{\partial y} = 3u^2 \frac{\partial u}{\partial y}

The inside is a single monomial plus a constant, so its partials are easy: ∂u∂x=2xy\frac{\partial u}{\partial x} = 2xy and ∂u∂y=x2\frac{\partial u}{\partial y} = x^2. The constant 11 contributes nothing to either.

∂f∂x=3(x2y+1)2⋅2xy=6xy(x2y+1)2\frac{\partial f}{\partial x} = 3\left(x^2y+1\right)^2 \cdot 2xy = 6xy\left(x^2y+1\right)^2
∂f∂y=3(x2y+1)2⋅x2\frac{\partial f}{\partial y} = 3\left(x^2y+1\right)^2 \cdot x^2

At (1,1)(1,1) the inside is 22, so the shared factor is 3⋅4=123 \cdot 4 = 12. That gives ∂f∂x=24\frac{\partial f}{\partial x} = 24 and ∂f∂y=12\frac{\partial f}{\partial y} = 12.

The mistake: expanding first, or forgetting the inner factor

Two failure modes dominate. The first is writing ∂f∂x=3(x2y+1)2\frac{\partial f}{\partial x} = 3\left(x^2y+1\right)^2 and stopping, which leaves out ∂u∂x=2xy\frac{\partial u}{\partial x} = 2xy. The chain rule is not optional once the inside depends on the variable.

The second is expanding the cube before differentiating. That is legal but wasteful: (x2y+1)3=x6y3+3x4y2+3x2y+1\left(x^2y+1\right)^3 = x^6y^3 + 3x^4y^2 + 3x^2y + 1, and differentiating gives 6x5y3+12x3y2+6xy6x^5y^3 + 12x^3y^2 + 6xy, which is the same as 6xy(x2y+1)26xy\left(x^2y+1\right)^2 in disguise.

  • Expanding costs time and creates more places to slip on an exponent.
  • The factored form makes it obvious where the partials vanish.
  • If you do expand, expand both answers before comparing, or they will look different when they are not.

Reading the answers

Both partials share the factor (x2y+1)2\left(x^2y+1\right)^2, which is zero exactly when x2y=−1x^2y = -1. Along that curve the gradient is the zero vector, so the surface is flat there in every direction.

Away from that curve, ∂f∂y\frac{\partial f}{\partial y} vanishes only when x=0x = 0, and ∂f∂x\frac{\partial f}{\partial x} vanishes when x=0x = 0 or y=0y = 0. So the whole yy axis is flat in both directions.

Notice that ∂f∂y\frac{\partial f}{\partial y} is never negative anywhere, since 3x23x^2 and the squared factor are both non-negative. It touches zero only on the yy axis and on the curve x2y=−1x^2y = -1, and never dips below, so the surface is non-decreasing in yy along every vertical line.

Frequently asked questions

Does the chain rule work the same way in two variables?

For an outer function of a single inner expression, yes. You differentiate the outer function as usual and multiply by the partial derivative of the inside with respect to your chosen variable, holding the other variable fixed throughout.

Why do both partials share the same squared factor?

Because both come from differentiating the same cube. The outer derivative 3u23u^2 depends only on u=x2y+1u = x^2y + 1, not on which variable you are moving, so it factors out of both answers and only the inner partial differs.