Partial Derivatives Basics: First-Order Derivatives & Rate-of-Change Interpretation

Geometrically, the partial derivative ∂z/∂x at a point (a, b) represents the slope of the tangent line to the curve formed by the intersection of the surface z = f(x, y) and the vertical plane y = b. This means we hold the y-coordinate constant at b and look at how z changes as we move only in the x-direction. Therefore, it measures the instantaneous rate of change of the function's output z with respect to a change in the input x, while the other input y is held fixed.

To find the derivative with respect to x, we must use the product rule because x appears in two places: as the first term 'x' and inside the natural log function. The product rule states that the derivative of u*v is u'v + uv'. Let u = x and v = ln(xy). The derivative of u with respect to x is 1. The derivative of v with respect to x requires the chain rule: the derivative of ln(xy) is 1/(xy) multiplied by the derivative of the inside (xy), which is y. So v' = (1/xy) * y = 1/x. Putting it all together: (1) * ln(xy) + (x) * (1/x) = ln(xy) + 1.

To find the partial derivative with respect to x, we treat y as a constant. The function is sin(3x + 2y). We use the chain rule: the derivative of sin(u) with respect to x is cos(u) times the partial derivative of u with respect to x. Here, u = 3x + 2y. First, compute ∂u/∂x: the derivative of 3x is 3, and the derivative of 2y (with respect to x) is 0. So ∂u/∂x = 3. Then, ∂z/∂x = cos(3x + 2y) * 3 = 3cos(3x + 2y).

The sign of a partial derivative indicates the direction of change. A positive partial derivative with respect to x means that if we increase x (move in the positive x-direction) while holding y constant, the function's value f will increase. A negative partial derivative with respect to y means that if we increase y (move in the positive y-direction) while holding x constant, the function's value f will decrease. Therefore, this combination describes how the function behaves with respect to small changes in each independent variable separately.

In this context, x represents the quantity of one product. The partial derivative ∂P/∂x measures how the profit P changes when we change only the quantity x of the first product, while keeping the quantity y of the second product fixed. The expression -2x gives this rate of change. Therefore, it represents the instantaneous rate of change of profit with respect to changes in the quantity of product x, assuming the quantity of product y does not change.

The partial derivative ∂T/∂x measures the instantaneous rate of change of the temperature T with respect to a change in the x-coordinate, while the y-coordinate is held constant. At the specific point (3,4), ∂T/∂x gives the rate at which temperature is changing per unit increase in x, if we were to move from (3,4) in the positive x-direction while staying at y=4. This is a direct interpretation of the partial derivative in a physical context.

A contour plot shows curves of constant function value. To estimate the partial derivative with respect to x at a point, we want to see how the function value changes as we move in the horizontal (x) direction. By moving horizontally from the point, we cross different contour lines. The rate at which the function value changes (the contour labels) per unit distance in the x-direction gives an estimate of ∂f/∂x. Specifically, if we move a small horizontal distance Δx and the function value changes by Δf (indicated by the contour lines crossed), then ∂f/∂x ≈ Δf/Δx.

To find the partial derivative with respect to x, we treat the variable y as a constant and differentiate the function with respect to x. The term 3x²y becomes 3(2x)(y) = 6xy, because the derivative of x² is 2x and the y is treated as a constant multiplier. The term 4y³ is treated as a constant since it has no x, and the derivative of a constant is 0. Therefore, the partial derivative is 6xy.

To find the partial derivative with respect to y, we treat the variable x as a constant. The function is a product of sin(2x) and  e3y. Since sin(2x) is treated as a constant coefficient, we only need to differentiate  e3y with respect to y. The derivative of  e3y  is  e3y multiplied by the derivative of the exponent (3y), which is 3. Therefore, the partial derivative is sin(2x) multiplied by 3e3y.

To find the partial derivative with respect to z, we treat the variables x and y as constants and differentiate the function with respect to z. The first term, xy, has no z and is therefore a constant, so its derivative is 0. The second term, yz², becomes y * 2z = 2yz. The third term, x*z, becomes x * 1 = x. Adding these results together gives 2yz + x.

To approximate the partial derivative with respect to x at (2, 3), we need to see how f changes when x changes while y is held constant at 3. Looking at the table, we have f(2, 3) = 10 and f(2.1, 3) = 10.5. The change in x (denoted Δx) is 2.1 - 2 = 0.1. The change in f (denoted Δf) is 10.5 - 10 = 0.5. The approximation for the partial derivative is the change in f divided by the change in x: Δf/Δx = 0.5 / 0.1 = 5.

To find the partial derivative with respect to y, f_y, we treat x as a constant and use the quotient rule. The quotient rule states that for a function u/v, the derivative is (u'v - uv') / v². Here, u = 5x - y⁴ and v = x + 2y. First, find u' (the derivative of u with respect to y): the derivative of 5x is 0, and the derivative of -y⁴ is -4y³, so u' = -4y³. Next, find v' (the derivative of v with respect to y): the derivative of x is 0, and the derivative of 2y is 2, so v' = 2. Applying the quotient rule: f_y = [ (-4y³)(x+2y) - (5x - y⁴)(2) ] / (x+2y)².

The partial derivative ∂C/∂x represents the marginal cost of producing good x, meaning the instantaneous rate of change of total cost with respect to the quantity x, while holding the quantity y fixed. To find its value, compute ∂C/∂x = 30 + 0.2x. Then evaluate at x=100: ∂C/∂x = 30 + 0.2*100 = 30 + 20 = 50. So the marginal cost is $50. Note that the value of y does not affect ∂C/∂x because there is no cross term involving x and y in the derivative.

To find the partial derivative with respect to y, we treat x as a constant. The derivative of ln(u) with respect to y is (1/u) * (∂u/∂y), where u = 3x+5y. First, compute ∂u/∂y: the derivative of 3x is 0, and the derivative of 5y is 5. So, ∂z/∂y = (1/(3x+5y)) * 5 = 5/(3x+5y).

First, find the general partial derivative ∂z/∂x. Treating y as a constant, the derivative of e^x * cos(y) is e^x * cos(y) (since cos(y) is a constant multiplier). The derivative of the term y is 0. So, ∂z/∂x = e^x * cos(y). Next, plug in the point (1, 0). We get e^1 * cos(0). Since e^1 = e and cos(0) = 1, the result is e * 1 = e.