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

1) What is the partial derivative of f(x, y) = 3x²y + 4y³ with respect to x?

6xy

6xy + 12y²

3x² + 12y²

3x² + 4y³

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.

Explanation

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.

2) For the function g(x, y) = sin(2x) * e^(3y), what is ∂g/∂y?

2cos(2x) * e^(3y)

Sin(2x) * 3e^(3y)

Cos(2x) * 3e^(3y)

Sin(2x) * e^(3y)

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.

Explanation

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.

3) What does the partial derivative ∂z/∂x represent geometrically for a surface z = f(x, y)?

Slope in the y-direction

Instantaneous rate of change of z as x changes while y is fixed

Curvature of the surface

Total rate of change in both variables

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.

Explanation

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.

4) Given h(x, y, z) = xy + yz² + xz, find ∂h/∂z.

Y + 2z + x

Y + 2yz + x

2yz + x

Y² + 2yz

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.

Explanation

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.

5) Calculate the partial derivative with respect to x for f(x, y) = x ln(xy).

Ln(xy) + 1

1/x

Ln(xy) + y

Y/x

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.

Explanation

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.

6) Consider a function f(x,y) such that f(2,3)=10, f(2.1,3)=10.5, and f(2,3.1)=10.2. Use a forward difference (with the next available x-value), approximate the partial derivative ∂f/∂x at the point (2, 3) using the given function values.

0.2

0.5

5

2

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.

Explanation

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.

7) Find ∂z/∂x for z = sin(3x + 2y).

3cos(3x + 2y)

2cos(3x + 2y)

Cos(3x + 2y)

3sin(3x + 2y)

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).

Explanation

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).

8) Find f_y for f(x, y) = (5x − y⁴)/(x + 2y).

(-4y³(x+2y) − (5x − y⁴)2)/(x+2y)²

(-4y³(x+2y) + (5x − y⁴)2)/(x+2y)²

(5(x+2y) − (5x − y⁴))/(x+2y)²

(-4y³(x+2y) − (5x − y⁴))/(x+2y)²

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.

Explanation

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.

9) At a point on a surface, the partial derivative ∂f/∂x is positive and ∂f/∂y is negative. What does this tell us?

Surface is flat

Moving +x increases f, moving +y decreases f

Moving +x decreases f, +y increases f

Function has maximum

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.

Explanation

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.

10) A company's cost C to produce two goods is given by C(x,y) = 500 + 30x + 20y + 0.1x² + 0.05y², where x and y are quantities. What does ∂C/∂x represent, and what is its value when x=100, y=200?

The marginal cost of producing good x when x=100 and y=200 is $50.

The marginal cost of producing good x when x=100 and y=200 is $30.

The marginal cost of producing good x when x=100 and y=200 is $70.

The marginal cost of producing good y when x=100 and y=200 is $20.

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.

Explanation

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.

11) If the function P(x, y) = 100 - x² - 2y² models the profit from selling two products, what does ∂P/∂x = -2x represent?

The rate at which profit changes as the price of product x changes.

The rate at which profit changes as the quantity of product x changes, holding y constant.

The total profit from selling product x.

The marginal profit of product y.

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.

Explanation

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.

12) Compute ∂z/∂y for the function z = ln(3x + 5y).

1/(3x+5y)

3/(3x+5y)

5/(3x+5y)

5/(3+5)

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).

Explanation

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).

13) Evaluate the partial derivative ∂z/∂x at the point (1, 0) for the function z = e^(x) * cos(y) + y.

0

1

E

E+1

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.

Explanation

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.

14) The temperature T at a point (x, y) on a metal plate is given by T(x, y) = 50 + x² - y². What is the meaning of ∂T/∂x at (3, 4)?

The rate of change of temperature as we move in the x-direction from the point (3,4), keeping y fixed.

The total temperature change in both directions.

The rate of change of temperature as we move in the y-direction from the point (3,4).

The temperature at the point (3,4).

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.

Explanation

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.

15) Given the contour plot of a function f(x, y), how can you estimate ∂f/∂x at a point?

By counting the number of contour lines crossing the y-axis.

By looking at how closely spaced the contour lines are in the y-direction.

By observing the change in the function's value as you move horizontally from the point, crossing contour lines.

By measuring the slope of the tangent line to the contour line at that point.

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.

Explanation

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.