Implicit Differentiation With Products & Mixed Terms

1) What does the chain rule tell us about differentiating terms with y when y is a function of x?

Terms with y have a derivative of 0

Terms with y keep their original coefficient only

We multiply by dy/dx when differentiating terms with y

We ignore terms with y in the differentiation process

When differentiating implicitly, we treat y as a function of x. Therefore, for any term involving y, we apply the chain rule. For example, d/dx(y²) = 2y(dy/dx) and d/dx(y³) = 3y²(dy/dx). The chain rule requires multiplying by the derivative of y with respect to x.

Explanation

When differentiating implicitly, we treat y as a function of x. Therefore, for any term involving y, we apply the chain rule. For example, d/dx(y²) = 2y(dy/dx) and d/dx(y³) = 3y²(dy/dx). The chain rule requires multiplying by the derivative of y with respect to x.

2) Find dy/dx for 2x³ + 3xy + y² = 10.

Dy/dx = -(6x² + 3y)/(3x + 2y)

Dy/dx = (6x² + 3y)/(3x + 2y)

Dy/dx = 6x² + 3y + 2y

Dy/dx = -(6x²)/(3x + 2y)

Differentiating both sides with respect to x: 6x² + [3y + 3x(dy/dx)] + 2y(dy/dx) = 0. Simplifying: 6x² + 3y + 3x(dy/dx) + 2y(dy/dx) = 0. Grouping dy/dx terms: 3x(dy/dx) + 2y(dy/dx) = -6x² - 3y. So (3x + 2y)(dy/dx) = -(6x² + 3y). Therefore: dy/dx = -(6x² + 3y)/(3x + 2y).

Explanation

Differentiating both sides with respect to x: 6x² + [3y + 3x(dy/dx)] + 2y(dy/dx) = 0. Simplifying: 6x² + 3y + 3x(dy/dx) + 2y(dy/dx) = 0. Grouping dy/dx terms: 3x(dy/dx) + 2y(dy/dx) = -6x² - 3y. So (3x + 2y)(dy/dx) = -(6x² + 3y). Therefore: dy/dx = -(6x² + 3y)/(3x + 2y).

3) Which of the following is the best definition of implicit differentiation?

Solving for y explicitly before differentiating

Differentiating both sides of an equation treating y as a function of x

Using only the power rule to find derivatives

Substituting numerical values before differentiation

Implicit differentiation is the technique of differentiating both sides of an equation with respect to x while treating y as an implicit function of x. This allows us to find dy/dx even when the equation cannot be solved explicitly for y in terms of x.

Explanation

Implicit differentiation is the technique of differentiating both sides of an equation with respect to x while treating y as an implicit function of x. This allows us to find dy/dx even when the equation cannot be solved explicitly for y in terms of x.

4) At the point (1, 1), what is the slope of the curve x³ + y³ = 2xy?

12/11

12/12

1/1

Undefined

We first find dy/dx by differentiating both sides: 3x² + 3y²(dy/dx) = 2y + 2x(dy/dx). Rearranging gives us 3x² - 2y = 2x(dy/dx) - 3y²(dy/dx) = (2x - 3y²)(dy/dx). So dy/dx = (3x² - 2y)/(2x - 3y²). At (1, 1), we plug-in the values to get dy/dx = (3(1)² - 2(1))/(2(1) - 3(1)²) = (3 - 2)/(2 - 3) = 1/(-1) = -1.

Explanation

We first find dy/dx by differentiating both sides: 3x² + 3y²(dy/dx) = 2y + 2x(dy/dx). Rearranging gives us 3x² - 2y = 2x(dy/dx) - 3y²(dy/dx) = (2x - 3y²)(dy/dx). So dy/dx = (3x² - 2y)/(2x - 3y²). At (1, 1), we plug-in the values to get dy/dx = (3(1)² - 2(1))/(2(1) - 3(1)²) = (3 - 2)/(2 - 3) = 1/(-1) = -1.

5) For y = x^y, what is the correct first step in finding dy/dx?

Take the natural logarithm of both sides

Write it as y = e^(y ln x)

Use the definition of exponents

Both A and B are correct first steps

For equations of the form y = x^y, we need to use logarithms before differentiating. We can either take ln of both sides: ln(y) = y ln(x), or rewrite as y = e^(y ln x). Both approaches allow us to apply implicit differentiation successfully.

Explanation

For equations of the form y = x^y, we need to use logarithms before differentiating. We can either take ln of both sides: ln(y) = y ln(x), or rewrite as y = e^(y ln x). Both approaches allow us to apply implicit differentiation successfully.

6) When finding dy/dx for x² + y² + 2x = 6, what is the derivative of 2x with respect to x?

2(dy/dx)

1/2

12/12

2x

The term 2x is a function of x only (not y), so its derivative with respect to x is simply 2. Only terms that explicitly contain y require us to apply the chain rule and multiply by dy/dx.

Explanation

The term 2x is a function of x only (not y), so its derivative with respect to x is simply 2. Only terms that explicitly contain y require us to apply the chain rule and multiply by dy/dx.

7) Find dy/dx for x²y + 3y = 8.

Dy/dx = -2xy/(x² + 3)

Dy/dx = 2xy/(x² + 3)

Dy/dx = 8/(x² + 3)

Dy/dx = -8/(x² + 3)

Differentiating: 2xy + x²(dy/dx) + 3(dy/dx) = 0. Grouping dy/dx terms: x²(dy/dx) + 3(dy/dx) = -2xy. So (x² + 3)(dy/dx) = -2xy. Therefore: dy/dx = -2xy/(x² + 3).

Explanation

Differentiating: 2xy + x²(dy/dx) + 3(dy/dx) = 0. Grouping dy/dx terms: x²(dy/dx) + 3(dy/dx) = -2xy. So (x² + 3)(dy/dx) = -2xy. Therefore: dy/dx = -2xy/(x² + 3).

8) For which type of functions is implicit differentiation particularly useful?

Only linear functions

Functions where y cannot be solved explicitly in terms of x

Only polynomial functions

Functions where x cannot be solved explicitly in terms of y

Implicit differentiation is especially valuable when we cannot or do not want to solve for y explicitly as a function of x. This includes cases where solving would be extremely difficult or where multiple y-values correspond to a single x-value (like circles and ellipses).

Explanation

Implicit differentiation is especially valuable when we cannot or do not want to solve for y explicitly as a function of x. This includes cases where solving would be extremely difficult or where multiple y-values correspond to a single x-value (like circles and ellipses).

9) Find dy/dx for (x² + y²)² = x² - y².

Dy/dx = (2x - 2x(x² + y²))/(2y(x² + y²) + 2y)

Dy/dx = (x(1-2(x² + y²)))/(y(1+2(x² + y²)))

Dy/dx = x/y

Dy/dx = y/x

Differentiating both sides: 2(x² + y²)(2x + 2y(dy/dx)) = 2x - 2y(dy/dx). Simplifying: 4x(x² + y²) + 4y(x² + y²)(dy/dx) = 2x - 2y(dy/dx). Collecting terms: 4y(x² + y²)(dy/dx) + 2y(dy/dx) = 2x - 4x(x² + y²). Factoring: (4y(x² + y²) + 2y)(dy/dx) = 2x(1 - 2(x² + y²)). So dy/dx = 2x(1 - 2(x² + y²))/[2y(2(x² + y²) + 1)] = x(1 - 2(x² + y²))/[y(2(x² + y²) + 1)].

Explanation

Differentiating both sides: 2(x² + y²)(2x + 2y(dy/dx)) = 2x - 2y(dy/dx). Simplifying: 4x(x² + y²) + 4y(x² + y²)(dy/dx) = 2x - 2y(dy/dx). Collecting terms: 4y(x² + y²)(dy/dx) + 2y(dy/dx) = 2x - 4x(x² + y²). Factoring: (4y(x² + y²) + 2y)(dy/dx) = 2x(1 - 2(x² + y²)). So dy/dx = 2x(1 - 2(x² + y²))/[2y(2(x² + y²) + 1)] = x(1 - 2(x² + y²))/[y(2(x² + y²) + 1)].

10) How many points on the curve x³ + y³ = 3xy have horizontal tangents (slope = 0)?

0

1

2

3

First, we find the derivative using implicit differentiation, which gives dy/dx = (y - x²)/(y² - x). A horizontal tangent occurs when the slope dy/dx is zero. This requires the numerator to be zero and the denominator to be non-zero. Setting the numerator to zero gives y - x² = 0, or y = x². We substitute this into the original equation: x³ + (x²)³ = 3x(x²), which simplifies to x⁶ - 2x³ = 0, or x³(x³ - 2) = 0. This yields two potential x-values: x = 0 and x = ∛2. If x = 0, then y = 0² = 0. The point is (0, 0). Let's check the denominator of dy/dx at this point: y² - x = 0² - 0 = 0. Since the denominator is zero, the slope is undefined, not zero. Thus, the tangent is not horizontal at the origin. On the other hand, if x = ∛2, then y = (∛2)² = ∛4. The point is (∛2, ∛4). The denominator at this point is (∛4)² - ∛2 = ∛16 - ∛2, which is not zero. Therefore, there is only one point where the tangent line is horizontal.

Explanation

First, we find the derivative using implicit differentiation, which gives dy/dx = (y - x²)/(y² - x). A horizontal tangent occurs when the slope dy/dx is zero. This requires the numerator to be zero and the denominator to be non-zero. Setting the numerator to zero gives y - x² = 0, or y = x². We substitute this into the original equation: x³ + (x²)³ = 3x(x²), which simplifies to x⁶ - 2x³ = 0, or x³(x³ - 2) = 0. This yields two potential x-values: x = 0 and x = ∛2. If x = 0, then y = 0² = 0. The point is (0, 0). Let's check the denominator of dy/dx at this point: y² - x = 0² - 0 = 0. Since the denominator is zero, the slope is undefined, not zero. Thus, the tangent is not horizontal at the origin. On the other hand, if x = ∛2, then y = (∛2)² = ∛4. The point is (∛2, ∛4). The denominator at this point is (∛4)² - ∛2 = ∛16 - ∛2, which is not zero. Therefore, there is only one point where the tangent line is horizontal.

11) Find dy/dx for x = y².

Dy/dx = 1/(2y)

Dy/dx = 2y

Dy/dx = y²/x

Dy/dx = x/2

Differentiating both sides with respect to x: 1 = 2y(dy/dx). Solving for dy/dx: dy/dx = 1/(2y).

Explanation

Differentiating both sides with respect to x: 1 = 2y(dy/dx). Solving for dy/dx: dy/dx = 1/(2y).

12) What is d²y/dx² at (1, 2) for the curve defined by y² = 2x + 2?

-1/8

1/8

-1/4

1/4

First find dy/dx: 2y(dy/dx) = 2, so dy/dx = 1/y. Now find d²y/dx²: d/dx[1/y] = -1/y²(dy/dx). Since dy/dx = 1/y, we get d²y/dx² = -1/y² · (1/y) = -1/y³. At (1, 2): d²y/dx² = -1/(2)³ = -1/8.

Explanation

First find dy/dx: 2y(dy/dx) = 2, so dy/dx = 1/y. Now find d²y/dx²: d/dx[1/y] = -1/y²(dy/dx). Since dy/dx = 1/y, we get d²y/dx² = -1/y² · (1/y) = -1/y³. At (1, 2): d²y/dx² = -1/(2)³ = -1/8.

13) When differentiating 3^(xy) with respect to x using implicit differentiation, what is the derivative?

3^(xy) ln 3

3^(xy) · y

3^(xy) ln 3 · (y + x dy/dx)

3^(xy) · x · y

Let u = xy. Then 3^(xy) = 3^u. The derivative of a^u with respect to x is a^u ln a · du/dx. Here, u = xy, so du/dx = y + x(dy/dx) using the product rule. Therefore, d/dx[3^(xy)] = 3^(xy) ln 3 · (y + x dy/dx).

Explanation

Let u = xy. Then 3^(xy) = 3^u. The derivative of a^u with respect to x is a^u ln a · du/dx. Here, u = xy, so du/dx = y + x(dy/dx) using the product rule. Therefore, d/dx[3^(xy)] = 3^(xy) ln 3 · (y + x dy/dx).

14) For the equation x cos y + y sin x = 1, what is dy/dx?

Dy/dx = (cos y + y cos x)/(x sin y - sin x)

Dy/dx =(sin x - cos y)/(x sin y + cos x)

Dy/dx = (cos x - sin y)/(sin y + x cos x)

Dy/dx = (sin x + cos y)/(sin y + x cos x)

Differentiating both sides: d/dx[x cos y] + d/dx[y sin x] = 0. For x cos y: cos y · 1 + x · (-sin y)(dy/dx) = cos y - x sin y(dy/dx). For y sin x: sin x · (dy/dx) + y · cos x. Combining: cos y - x sin y(dy/dx) + sin x(dy/dx) + y cos x = 0. Collecting dy/dx terms: sin x(dy/dx) - x sin y(dy/dx) = -cos y - y cos x. So (sin x - x sin y)(dy/dx) = -(cos y + y cos x). Therefore: dy/dx = (cos y + y cos x)/(x sin y - sin x).

Explanation

Differentiating both sides: d/dx[x cos y] + d/dx[y sin x] = 0. For x cos y: cos y · 1 + x · (-sin y)(dy/dx) = cos y - x sin y(dy/dx). For y sin x: sin x · (dy/dx) + y · cos x. Combining: cos y - x sin y(dy/dx) + sin x(dy/dx) + y cos x = 0. Collecting dy/dx terms: sin x(dy/dx) - x sin y(dy/dx) = -cos y - y cos x. So (sin x - x sin y)(dy/dx) = -(cos y + y cos x). Therefore: dy/dx = (cos y + y cos x)/(x sin y - sin x).

15) A particle moves along a path defined by x² + xy + y² = 7. At the point (2, 1), what is the instantaneous rate of change of y with respect to x?

-5/4

-4/5

4/5

5/4

This is asking for dy/dx at the point (2, 1). First verify that (2, 1) is on the curve: 2² + 2(1) + 1² = 4 + 2 + 1 = 7, which is correct. Now find dy/dx: Differentiating both sides: 2x + [y + x(dy/dx)] + 2y(dy/dx) = 0. So 2x + y + x(dy/dx) + 2y(dy/dx) = 0. Grouping dy/dx terms: x(dy/dx) + 2y(dy/dx) = -2x - y. So (x + 2y)(dy/dx) = -(2x + y). Therefore: dy/dx = -(2x + y)/(x + 2y). At (2, 1): dy/dx = -(2(2) + 1)/(2 + 2(1)) = -(4 + 1)/(2 + 2) = -5/4.

Explanation

This is asking for dy/dx at the point (2, 1). First verify that (2, 1) is on the curve: 2² + 2(1) + 1² = 4 + 2 + 1 = 7, which is correct. Now find dy/dx: Differentiating both sides: 2x + [y + x(dy/dx)] + 2y(dy/dx) = 0. So 2x + y + x(dy/dx) + 2y(dy/dx) = 0. Grouping dy/dx terms: x(dy/dx) + 2y(dy/dx) = -2x - y. So (x + 2y)(dy/dx) = -(2x + y). Therefore: dy/dx = -(2x + y)/(x + 2y). At (2, 1): dy/dx = -(2(2) + 1)/(2 + 2(1)) = -(4 + 1)/(2 + 2) = -5/4.

Implicit Differentiation with Products & Mixed Terms