Find Function From Partial Derivatives Calculator

Determine if a potential function f(x,y) exists for a given vector field by checking the exactness condition (Clairaut’s Theorem).

Define Partial Derivatives (Linear Forms)

Input coefficients for the partial derivatives ∂f/∂x = P(x,y) and ∂f/∂y = Q(x,y).
We assume linear forms: P = Ax + By + C and Q = Dx + Ey + F.

Partial w.r.t x: P(x,y) = Ax + By + C

Partial w.r.t y: Q(x,y) = Dx + Ey + F

Visualization Domain

Exactness Analysis Results

Based on Clairaut’s Theorem, since the mixed partial derivatives are equal (∂P/∂y = ∂Q/∂x), the vector field is conservative and a potential function f(x,y) can be found.

Comparison of Mixed Partial Derivatives for Exactness Test

Derivative Type	Mathematical Notation	Calculated Value
Partial of P w.r.t y	∂/∂y(Ax + By + C)	3
Partial of Q w.r.t x	∂/∂x(Dx + Ey + F)	3

What is a “Find Function from Partial Derivatives Calculator”?

In multivariable calculus, a common problem is the reverse process of differentiation: given the partial derivatives of an unknown function, can we recover the original function? A “find function from partial derivatives calculator” typically helps solve problems where you are given two functions, P(x,y) and Q(x,y), and asked to find a function f(x,y) such that:

• ∂f/∂x = P(x,y)
• ∂f/∂y = Q(x,y)

This process is analogous to finding the antiderivative in single-variable calculus. In physics, this is known as finding the **potential function** f for a given **vector field** F = ⟨P, Q⟩. If such a function exists, the vector field is called a **conservative vector field**.

Students in calculus III, physics, or engineering often use tools to verify if a function can be found before attempting the sometimes lengthy integration process. A common misconception is that *any* pair of partial derivatives P and Q corresponds to a single function f. This is not true; a specific condition must be met.

The Mathematical Formula: Clairaut’s Theorem and Integration

To find the function from partial derivatives, we must first establish that such a function actually exists. We rely on **Clairaut’s Theorem** on the equality of mixed partial derivatives.

Step 1: The Exactness Test (Clairaut’s Theorem)

For a function f(x,y) to exist that satisfies the given partials, the mixed second-order partial derivatives must be equal (assuming continuity). This provides a necessary and sufficient condition:

∂P/∂y = ∂Q/∂x

If this condition holds, the expression P dx + Q dy is called an **exact differential**, and the function f(x,y) exists. If they are not equal, no such function f exists.

Step 2: The Integration Procedure

If the test passes, we find f(x,y) through a two-step integration process:

1. Integrate P(x,y) with respect to x. The “constant of integration” here is actually an arbitrary function of y, denoted g(y):
   
   f(x,y) = ∫ P(x,y) dx + g(y)
2. Differentiate this result with respect to y, and set it equal to the given Q(x,y):
   
   ∂/∂y [∫ P(x,y) dx] + g'(y) = Q(x,y)
3. Solve the resulting equation for g'(y), then integrate with respect to y to find g(y).
4. Substitute g(y) back into the equation from step 1 to get the final f(x,y) + C, where C is a true constant.

Variable Definitions in Multivariable Integration

Variable/Symbol	Meaning	Typical Context
f(x,y)	The unknown potential function we want to find.	Scalar Field
P(x,y) or M	The given partial derivative with respect to x (∂f/∂x).	x-component of Vector Field
Q(x,y) or N	The given partial derivative with respect to y (∂f/∂y).	y-component of Vector Field
∂P/∂y	The mixed partial derivative of P w.r.t y.	Test condition
g(y) or h(x)	The “constant” of integration function.	Intermediate step

Practical Examples: Finding the Function

Example 1: A Conservative Field (Function Exists)

Problem: Find f(x,y) given ∂f/∂x = 2xy and ∂f/∂y = x² + 3y².

• Here, P = 2xy and Q = x² + 3y².
• Test: ∂P/∂y = 2x. ∂Q/∂x = 2x. They are equal. The function exists.
• Integrate P w.r.t x: f(x,y) = ∫(2xy)dx = x²y + g(y).
• Differentiate w.r.t y: ∂f/∂y = ∂/∂y(x²y + g(y)) = x² + g'(y).
• Set equal to Q: x² + g'(y) = x² + 3y². This simplifies to g'(y) = 3y².
• Find g(y): g(y) = ∫(3y²)dy = y³.
• Final Answer: f(x,y) = x²y + y³ + C.

Example 2: A Non-Conservative Field (No Function Exists)

Problem: Find f(x,y) given ∂f/∂x = y and ∂f/∂y = -x.

• Here, P = y and Q = -x.
• Test: Calculate the mixed partials.
• ∂P/∂y = ∂/∂y(y) = 1.
• ∂Q/∂x = ∂/∂x(-x) = -1.
• Conclusion: Since 1 ≠ -1, the condition fails. No function f(x,y) exists that satisfies these partial derivatives. The vector field is not conservative (it has “curl”).

How to Use This Exactness Calculator

This calculator specifically handles the crucial first step: determining if a function *can* be found from given partial derivatives of linear forms. It visualizes the resulting vector field.

1. Identify P and Q: Look at your problem. The term representing ∂f/∂x is your P, and ∂f/∂y is your Q.
2. Input Coefficients: The calculator assumes forms P = Ax + By + C and Q = Dx + Ey + F. Enter the values for A, B, C, D, E, and F based on your problem.
3. Check the Result: The calculator immediately computes ∂P/∂y (which is B) and ∂Q/∂x (which is D).
4. Interpret Status: If the status is “Conservative,” the mixed partials match, and a function f(x,y) exists. If it says “Not Conservative,” no such function exists.
5. Visualize: The chart shows the vector field ⟨P,Q⟩. A conservative field (where the function exists) often looks like flow directly uphill or downhill without swirling circulation.

Key Factors Affecting the Existence of the Function

When trying to find a function from its partial derivatives, several mathematical factors come into play.

• Equality of Mixed Partials: As discussed, P_y = Q_x is the definitive test. If this fails, you cannot find the function.
• Domain Topology (Simply Connected): The test P_y = Q_x guarantees a potential function exists only on a “simply connected” domain (a region with no holes). If the domain has holes (e.g., the field is undefined at the origin), a function might not exist globally even if the mixed partials match locally.
• Continuity: The partial derivatives P and Q, and their first derivatives, must be continuous over the region of interest for Clairaut’s theorem to apply reliably.
• Path Independence: If the function f exists, the line integral of the vector field ∫ F · dr between any two points is independent of the path taken. This is a physical consequence of being a conservative field.
• Curl of the Field: In 3D calculus, the condition P_y = Q_x (along with corresponding checks for z-components) is equivalent to saying the **curl** of the vector field is zero (∇ × F = 0).
• The Integration Constant (+C): When you successfully find the function, it is always unique only up to an arbitrary additive constant C. f(x,y) and f(x,y) + 5 have the exact same partial derivatives.

Frequently Asked Questions (FAQ)

What if the mixed partial derivatives are not equal?

If ∂P/∂y ≠ ∂Q/∂x, it is impossible to find a single function f(x,y) that satisfies both given partial derivatives. The vector field is non-conservative.

Why do we add g(y) instead of just C when integrating w.r.t x?

When integrating with respect to x in multivariable calculus, y is treated as a constant. Therefore, the “constant of integration” can be any arbitrary function that depends only on y, because the derivative of g(y) with respect to x is zero.

Can this calculator solve for non-linear P and Q functions?

This specific calculator tool implements the exactness check for linear forms of P and Q (Ax+By+C). While the mathematical principle applies to non-linear functions (like P=y cos(x)), this input interface is restricted to linear coefficients for simplicity.

What is the physical meaning of finding this function?

In physics, if F = ⟨P,Q⟩ represents a force field (like gravity or electrostatics), finding the function f means finding the **potential energy function**. The force is the negative gradient of the potential (F = -∇f).

What is an “exact differential equation”?

An exact differential equation is of the form P(x,y)dx + Q(x,y)dy = 0, where the condition P_y = Q_x holds. The general solution is found by finding the function f(x,y) and setting it equal to a constant: f(x,y) = C.

Does this apply to functions of three variables f(x,y,z)?

Yes, but the test is more involved. You must check three pairs of mixed partials: ∂P/∂y = ∂Q/∂x, ∂P/∂z = ∂R/∂x, and ∂Q/∂z = ∂R/∂y, where the vector field is ⟨P, Q, R⟩.

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