Ch. 3 - Derivatives

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

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.6.78

Chapter 3, Problem 3.6.78

Find dy/dt when x = 1 if y = x² + 7x − 5 and dx/dt = ¹/₃.

Verified step by step guidance

First, identify the given function y = x² + 7x − 5. This is the function for which we need to find the derivative with respect to time, t.

To find dy/dt, use the chain rule. The chain rule states that dy/dt = (dy/dx) * (dx/dt). We need to find dy/dx first.

Differentiate y = x² + 7x − 5 with respect to x to find dy/dx. The derivative of x² is 2x, and the derivative of 7x is 7. Therefore, dy/dx = 2x + 7.

Substitute x = 1 into the expression for dy/dx to find the specific value of the derivative at x = 1. This gives dy/dx = 2(1) + 7.

Now, substitute dy/dx and dx/dt = ¹/₃ into the chain rule formula: dy/dt = (dy/dx) * (dx/dt). Calculate dy/dt using the values obtained in the previous steps.

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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 context, it helps us differentiate y with respect to t by considering the relationship between x and y, and using the chain rule to account for dx/dt.

Chain Rule

The chain rule is a fundamental concept 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. Here, it allows us to find dy/dt by multiplying dy/dx by dx/dt.

Substitution

Substitution involves replacing variables with known values to simplify expressions or solve equations. In this problem, after finding the expression for dy/dt, we substitute x = 1 and dx/dt = 1/3 to calculate the specific value of dy/dt at that point, providing a concrete solution to the problem.

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Related Practice

Textbook Question

Find an equation of the straight line having slope 1/4 that is tangent to the curve y = √x.

Textbook Question

Theory and Examples

Intersecting normal line The line that is normal to the curve x² + 2xy – 3y² = 0 at (1,1) intersects the curve at what other point?

Textbook Question

In Exercises 41–58, find dy/dt.

y = √(3t + (√2 + √(1 − t)))

Textbook Question

Finding Derivative Functions and Values

Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.

g(t) = 1/t²; g′(−1), g′(2), g′(√3)

Textbook Question

A growing sand pile Sand falls from a conveyor belt at the rate of 10 m³/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute.

Textbook Question

If r + s² + v³ = 12, dr/dt = 4, and ds/dt = –3, find dv/dt when r = 3 and s = 1.