Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.8.8

Chapter 3, Problem 3.8.8

If x²y³ = 4/27 and dy/dt = ¹/₂, then what is dx/dt when x = 2?

Verified step by step guidance

First, recognize that the given equation x²y³ = 4/27 is an implicit function of x and y. We need to differentiate both sides of this equation with respect to time t to find the relationship between dx/dt and dy/dt.

Apply implicit differentiation to the equation x²y³ = 4/27 with respect to t. Use the product rule and chain rule: d/dt(x²y³) = d/dt(4/27). Since 4/27 is a constant, its derivative is 0.

Differentiate the left side: d/dt(x²y³) = 2x(dx/dt)y³ + x²(3y²)(dy/dt). This uses the product rule: d(uv)/dt = u(dv/dt) + v(du/dt), where u = x² and v = y³.

Substitute the known values into the differentiated equation: x = 2, dy/dt = 1/2, and solve for dx/dt. The equation becomes: 2(2)(dx/dt)y³ + (2)²(3y²)(1/2) = 0.

Solve the resulting equation for dx/dt. Rearrange the terms to isolate dx/dt and substitute any additional known values to find the expression for dx/dt.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. In this problem, the equation x²y³ = 4/27 involves both x and y, so we differentiate both sides with respect to time t, applying the chain rule to account for the derivatives of x and y.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. In this context, it helps differentiate terms like x² and y³ with respect to t, considering dx/dt and dy/dt.

Solving for a Derivative

Once the implicit differentiation is performed, the next step is to solve for the desired derivative, dx/dt. This involves substituting known values, such as x = 2 and dy/dt = 1/2, into the differentiated equation and isolating dx/dt. This process requires algebraic manipulation to express dx/dt in terms of the given quantities.

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Find all points on the curve y = tan x, −π/2 < x < π/2, where the tangent line is parallel to the line y = 2x. Sketch the curve and tangent lines together, labeling each with its equation.

Textbook Question

Derivatives in Differential Form

In Exercises 17–28, find dy.

y = cos(x²)

Textbook Question

Approximation Error

In Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Find

a. the change Δf = f(x₀ + dx) − f(x₀);

b. the value of the estimate df = fʹ(x₀) dx; and

c. the approximation error |Δf − df|.

f(x) = x⁴, x₀ = 1, dx = 0.1

Textbook Question

Derivative of multiples Does knowing that a function g(t) is differentiable at t = 7 tell you anything about the differentiability of the function 3g at t = 7? Give reasons for your answer.

Textbook Question

Find the derivatives of the functions in Exercises 19–40.

y = (4x + 3)⁴(x + 1)⁻³

Textbook Question

Find the derivatives of the functions in Exercises 1–42.

𝔂 = 2 tan² x - sec² x