Partial Derivatives Made Simple: Definitions, Tables & Mixed Partials

The partial derivative with respect to y measures the rate of change as y varies while x remains constant. Therefore, the limit must show a change of h in the y-argument only.

The partial derivative of the production function P with respect to Labor (L) measures the instantaneous rate of change of production as Labor increases, while Capital (K) is held constant. In economics, this specific rate of change is defined as the marginal productivity of labor. It estimates the extra output generated by adding one more unit of labor.

We use the chain rule (general power rule). The outer function is u1/2, and the inner function is u = x² + y². The derivative of the outer function is (1/2)u-1/2. The partial derivative of the inner function with respect to x is 2x. Multiplying these gives: (1/2)(x² + y²)-1/2 * (2x). The 2s cancel out, leaving x * (x² + y²)-1/2, which is equivalent to x / (x² + y²)1/2.

To find the second partial derivative ∂²f/∂x², we first find ∂f/∂x, then differentiate that result again with respect to x. Step 1: Find ∂f/∂x. For f(x,y) = e^(x²y), we treat y as constant and use the chain rule. The derivative is e^(x²y) * ∂/∂x(x²y) = e^(x²y) * 2xy = 2xye^(x²y). Step 2: Differentiate ∂f/∂x with respect to x again. We have ∂f/∂x = 2xye^(x²y). Using the product rule (since we have 2xy times e^(x²y)), we get ∂²f/∂x² = ∂/∂x(2xy)e^(x²y) + 2xy∂/∂x(e^(x²y)). The first term is 2ye^(x²y) because ∂/∂x(2xy) = 2y (y constant). The second term is 2xy * e^(x²y) * 2xy = 2xy * 2xy * e^(x²y) = 4x²y²e^(x²y). Adding both terms gives 2ye^(x²y) + 4x²y²e^(x²*y). Option C only gives the first term, missing the second part from the product rule.

Clairaut's Theorem (also called the equality of mixed partials theorem) states that if a function f(x,y) has mixed partial derivatives fxy and fyx that are both continuous on an open region, then these mixed partial derivatives are equal at every point in that region. That is, fxy = fyx  everywhere. This tells us that for well-behaved functions, the order of differentiation does not matter. Option A is too weak (it says "some point" but the theorem guarantees equality everywhere). Option C contradicts the theorem. Option D is incorrect; Clairaut's theorem applies to many non-linear functions as long as the mixed partials are continuous.

To compute ∂²f/∂y∂x, we first differentiate with respect to x, then differentiate that result with respect to y. Differentiating w.r.t. x gives ∂f/∂x = 3x²y² + y. Differentiate this result w.r.t. y gives ∂/∂y (3x²y² + y) = 3x²(2y) + 1 = 6x²y + 1.

In a second-order linear PDE, the term ∂²f/∂x² represents the second partial derivative with respect to x. Geometrically, this tells us about the concavity or curvature of the function in the x-direction. More specifically, it measures how the slope in the x-direction (which is ∂f/∂x) changes as we move in the x-direction. This is analogous to the second derivative in single-variable calculus, which measures how the slope changes. Option B is incorrect because ∂²f/∂x² involves only x, not y. Option C confuses x and y directions. Option D incorrectly describes an average rather than an instantaneous rate of change of the slope.

The notation f_xyx means we differentiate with respect to x, then y, then x again. Differentiate w.r.t x: f_x = 4x³ y². Differentiate that result w.r.t y: f_xy = 4x³ * (2y) = 8x³ y. Differentiate that result w.r.t x: f_xyx = 8 * (3x²) * y = 24x² y.

To find the partial derivative ∂f/∂x, we treat y as a constant and differentiate with respect to x. Starting with f(x,y) = 3x²y + 5xy³, we handle each term separately. For the first term 3x²y, the derivative is 3y2x = 6xy because y is constant. For the second term 5xy³, the derivative is 5y³1 = 5y³ because y³ is constant. Adding these results gives ∂f/∂x = 6xy + 5y³. The other options incorrectly treat x as constant or mess up the coefficients.

The partial derivative ∂f/∂x is defined as the rate of change of the function f in the x-direction while treating all other variables (in this case y) as constants. At point (a,b), we imagine moving only along the x-axis, changing x slightly while keeping y fixed at its value b. This is fundamentally different from a total derivative which would allow y to change as x changes. Option B describes ∂f/∂y, not ∂f/∂x. Option C is vague and option D describes an average, not the precise instantaneous rate that a derivative measures.

When computing ∂f/∂y, we treat x and z as constants and differentiate with respect to y. The function is f(x,y,z) = x²y³z⁴. Since x² and z⁴ are constant factors, we focus on differentiating y³. Using the power rule, the derivative of y³ with respect to y is 3y². Therefore, ∂f/∂y = x²3y²z⁴ = 3x²y²z⁴. Option B incorrectly differentiates x² instead of y³. Option C differentiates with respect to z. Option D has an extra factor of 2 that shouldn't be there.

To find ∂f/∂x for f(x,y) = ln(x² + y²), we use the chain rule. The outer function is ln(u) and the inner function is u = x² + y². The derivative of ln(u) with respect to x is (1/u)*∂u/∂x. Here, u = x² + y², so ∂u/∂x = 2x (since y is constant). Therefore, ∂f/∂x = (1/(x² + y²))*2x = 2x/(x² + y²). Option B gives ∂f/∂y instead of ∂f/∂x. Option C forgets the chain rule factor. Option D incorrectly introduces a y factor.

To compute ∂f/∂y for f(x,y,z) = xyz + x²z, we treat x and z as constants and differentiate with respect to y. Looking at the first term xyz, the derivative with respect to y is xz (since ∂/∂y(y) = 1 and xz are constant factors). Looking at the second term x²z, notice that y does not appear at all in this term. When we treat x and z as constants, x²*z is a constant with respect to y, and its derivative is 0. Therefore, ∂f/∂y = xz + 0 = xz. Option B incorrectly tries to differentiate the second term and confuses variables. Option C forgets the x factor. Option D adds an unnecessary term.

To approximate fy(2,1), the partial derivative with respect to y at point (2,1), we use a difference quotient in the y-direction. We need values at y=0 and y=2 for x=2. The formula is fy(2,1) ≈ [f(2,2) - f(2,0)]/(2-0). From the table, f(2,2) = 12 and f(2,0) = 6. The numerator is 12 - 6 = 6. The denominator is 2. Therefore, fy(2,1) ≈ 6/2 = 3. This gives us the average rate of change in the y-direction at the point (2,1). Option A would be too small, option B would be [f(2,1)-f(2,0)]/1 = (8-6)/1 = 2 but uses less data, and option D is too large. The centered difference using points on both sides gives the most reliable approximation