Partial Derivatives
Differentiate with respect to one variable while holding the others fixed. Interpret coordinate slopes and compare mixed second partials, which agree when both mixed derivatives are continuous near the point.
What a learner can do afterwards
- Compute both first partials of a function of two variables
- Interpret a partial derivative as a slope along a coordinate direction
- Compute mixed second partials in both orders and check their agreement under continuity conditions
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For f(x, y), fx means differentiate in x with y fixed. For a positive integer n, the x derivative of x^n is n x^(n-1). Keep fixed factors as coefficients. Constants have derivative 0, so y-only terms contribute 0 to fx. For h = 2x² + 5xy + y³, hx = 4x + 5y, so hx(1, 2) = 14. You may substitute the fixed y-value early, but keep x variable until you have differentiated.
To find fy, hold x fixed and differentiate in y. Terms containing only x have derivative 0 in y. For h = 2x² + 5xy + y³, hy = 5x + 3y², so hy(1, 2) = 17. A fixed x-factor acts as a coefficient: the y derivative of x²y³ is 3x²y². Differentiate before substituting the requested y-value.
On z = f(x, y), fixing y = b leaves the trace z = f(x, b). Its tangent slope at x = a is fx(a, b): change in z per change in x. Fixing x = a instead leaves g(y) = f(a, y), whose tangent slope at y = b is g'(b) = fy(a, b). Positive and negative slopes mean that the tangent line rises or falls in the direction of the increasing input.
A second partial differentiates a first partial. Here fxy means differentiate in x, then y; fyx means y, then x. Keep the first partial as a function until both differentiations are finished. For h = 3x²y², hx = 6xy² and hxy = 12xy. In the other order, hy = 6x²y and hyx = 12xy. Evaluate at the point after finding the mixed derivative.
If fxy and fyx exist and are continuous on an open neighborhood of a point, they agree there. A smooth function has continuous derivatives of every order on an open domain, so its mixed partials agree at every point of that domain. A mismatch means at least one first or second differentiation needs checking. Agreement alone does not prove the first partials are correct: errors can disappear during the second differentiation.
A zero partial derivative at one point does not make the whole trace constant. For h = x² + y², hx(0, 2) = 0. Yet the trace h(x, 2) = x² + 4 gives 4 at x = 0 and 5 at x = 1. Its tangent is horizontal at x = 0, but its values are not constant.
The fixed-variable rule also applies to trigonometric terms. Use radians: the y derivative of sin y is cos y. With x fixed, the y derivative of x sin y is x cos y. For h = x sin y + 2y², hy = x cos y + 4y. At (3, 0), hy = 3 because cos 0 = 1.
Hold the other input fixed, differentiate, then evaluate. Continuous mixed partials agree, but agreement alone does not prove that every step is correct.
Lesson, questions and flashcards written for this learning dataset. OpenStax Calculus Volume 3 and Khan Academy materials were studied for mathematical facts and teaching structure.
Readings, practice questions and revised flashcards written for this learning dataset. OpenStax and Khan Academy materials informed factual study. The linked videos are by Khan Academy; their licenses remain unverified, and no CC0 grant is claimed.
More to read
Differentiate with one input fixed
A function f(x, y) has two inputs. To study its change in x, choose a y-value and keep it fixed. The remaining expression is an ordinary function of x. Its derivative gives fx at that fixed y-value.
For p(x, y) = 4x²y + 6y², treating y as a constant gives px = 8xy. The term 6y² contributes zero because it does not change with x. Holding x fixed instead gives py = 4x² + 12y. The same term can be constant in one calculation and variable in the other.
To evaluate a partial derivative at a point, differentiate and then insert the inputs. You may insert the held input early. Keep the input being differentiated variable until its derivative has been found. Substituting both inputs first produces a number and loses the variation you need to measure.
Read a slope from a coordinate trace
The graph z = f(x, y) is a surface. Fixing y = b restricts it to the trace z = f(x, b). Regard z as a function of x on this trace. At x = a, its tangent slope is fx(a, b). The slope compares change in height z with change in x.
Fixing x = a gives another trace, z = f(a, y). Its tangent slope at y = b is fy(a, b), measured with respect to y. These slopes can have different values and signs at the same surface point.
For s(x, y) = x² - 2y, the x trace at y = 1 is z = x² - 2. At x = 3 its slope is 6. The y trace at x = 3 is z = 9 - 2y, whose slope is -2. The first tangent rises in the increasing-x direction; the second falls in the increasing-y direction. A zero tangent slope at one point does not establish that the whole trace is constant.
What matching mixed partials establish
In the notation used here, fxy means take the x derivative first and then differentiate that result in y. For fyx, reverse those steps. Keep each first partial as a function while taking the second derivative. Evaluate at the requested point afterward.
For r(x, y) = 2x³y², rx = 6x²y² and rxy = 12x²y. Starting in y gives ry = 4x³y and ryx = 12x²y. Both routes give the same formula.
If both mixed derivatives exist and are continuous on an open neighborhood of a point, they agree there. Smooth functions satisfy this condition on their open domain. Without that continuity condition, agreement cannot be assumed for every function.
A mismatch under these conditions calls for checking both differentiation stages. Agreement alone does not prove the first partials are correct. For r = x³ + y³, proposed first partials 4x² and 4y² have matching mixed derivatives of zero. Yet direct differentiation gives 3x² and 3y². The mixed check misses those coefficient errors.
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