Partial Derivative of x^2y^3 + e^(xy): Worked Example

The problem: find ∂f/∂x for f(x, y) = x²y³ + e^(xy).

Students find this problem hard for two reasons that stack on top of each other. First, you have to treat y as a constant, which feels unnatural the first few times: your hand wants to differentiate every letter it sees. Second, the e^(xy) term needs the chain rule with a twist: the derivative of the inside function xy with respect to x is y, not 1, and forgetting that single letter quietly breaks the whole answer. This worked example handles both, one move at a time.

Worked solution

We want ∂f/∂x, the partial derivative with respect to x.

Step 1. State the rule you will use. The symbol ∂/∂x means: differentiate with respect to x while treating every other variable as a constant. Here, y behaves exactly like the number 5 would. Nothing about y changes when x changes, so anything built only from y is a constant factor.

Step 2. Split the sum. Derivatives act term by term, and partial derivatives are no exception:

∂/∂x [x²y³ + e^(xy)] = ∂/∂x (x²y³) + ∂/∂x (e^(xy))

Step 3. Differentiate the first term. In x²y³, the factor y³ is constant with respect to x, so it rides along untouched while x² differentiates normally:

∂/∂x (x²y³) = y³ · ∂/∂x (x²) = y³ · 2x = 2xy³ ✓

Step 4. Set up the chain rule for the second term. The term e^(xy) is a composition: the exponential function wrapped around the product xy. The chain rule says:

∂/∂x e^(u) = e^(u) · ∂u/∂x, where u = xy

In words: derivative of the outside (which for e^u is just e^u again) times the derivative of the inside.

Step 5. Find the inside derivative. This is the step most students get wrong. With y held constant:

∂/∂x (xy) = y · ∂/∂x (x) = y · 1 = y

Not 1. The y survives because it is a constant factor, and constant factors survive differentiation.

Step 6. Assemble the second term. Plug the inside derivative back into the chain rule:

∂/∂x (e^(xy)) = e^(xy) · y = y·e^(xy) ✓

Step 7. Add the pieces. The partial derivative is the sum of the two differentiated terms:

∂f/∂x = 2xy³ + y·e^(xy)

Bonus: ∂f/∂y. The same ideas with the roles swapped. Now x is the constant:

∂/∂y (x²y³) = x² · 3y² = 3x²y²
∂/∂y (e^(xy)) = e^(xy) · ∂/∂y (xy) = e^(xy) · x = x·e^(xy)

So ∂f/∂y = 3x²y² + x·e^(xy). Notice the symmetry: each partial derivative picks up the “other” variable as its chain-rule factor.

Evaluate at (1, 0): ∂f/∂x (1, 0) = 2(1)(0)³ + (0)·e^((1)(0)) = 0 + 0·1 = 0.

Why this method works

A partial derivative measures slope in exactly one direction: change x a little, hold y fixed, and see how f responds. Treating y as a constant is not a shortcut or a trick; it is the definition of “change x and nothing else.” The chain rule applies to e^(xy) because it is genuinely a composition of two functions, and the chain rule is the tool that differentiates compositions. Each step follows from a definition, not from memorized magic.

Common mistakes

Differentiating y as if it were x. Writing something like ∂/∂x (x²y³) = 2xy² · 3y is wrong twice over. With respect to x, y³ is a constant: its derivative is 0, and it stays glued to the term as a plain factor.
Chain-rule amnesia. Answering e^(xy) for the second term and stopping. The exponent xy depends on x, so the inside derivative y must multiply the result. Forgetting it is the single most common point loss on this problem type.
Writing the inside derivative as 1. ∂/∂x (xy) = y, not 1 and not 0. The product xy has two factors; one of them (x) differentiates to 1 while the other (y) stays, leaving y.

Check (never skip)

Verify with a finite-difference test at (1, 1). Our formula gives ∂f/∂x (1, 1) = 2(1)(1)³ + (1)·e^1 = 2 + e ≈ 4.7183. Now approximate numerically: f(1.001, 1) = (1.001)²(1)³ + e^(1.001) ≈ 1.002001 + 2.721001 = 3.723002, and f(1, 1) = 1 + e ≈ 3.718282. Then [f(1.001, 1) − f(1, 1)] / 0.001 ≈ 0.004720 / 0.001 = 4.720, which matches 2 + e ≈ 4.7183 to the accuracy a 0.001 step allows. ✓ The formula survives contact with actual numbers.

Frequently Asked Questions

Q: What does ∂f/∂x actually mean?
A: It is the rate at which f changes when you nudge x a tiny amount while holding y fixed. Geometrically, it is the slope of the surface z = f(x, y) in the x-direction at that point.

Q: Why is the derivative of xy with respect to x equal to y?
A: Because y is treated as a constant, and the derivative of (constant)·x is the constant itself. Formally: ∂/∂x (xy) = y · ∂/∂x (x) = y · 1 = y.

Q: Do I always need the chain rule when I see e raised to something?
A: Yes, whenever the exponent depends on the variable you are differentiating with respect to. Only the bare e^x differentiates to itself with no extra factor; e^(xy), e^(x²), e^(sin x) all need the chain rule.

Try it with steps on our partial derivative calculator.