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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.87
Chapter 3, Problem 3.87

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.
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x + √xy = 6, (4, 1)

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First, understand that the problem requires finding the tangent and normal lines to the curve given by the equation \(x + \sqrt{xy} = 6\) at the point (4, 1). The tangent line is the line that just touches the curve at this point, while the normal line is perpendicular to the tangent line at this point.
To find the equation of the tangent line, we need to determine the derivative of the curve with respect to x, which will give us the slope of the tangent line at the point (4, 1). Start by differentiating both sides of the equation \(x + \sqrt{xy} = 6\) with respect to x. Use implicit differentiation since y is a function of x.
Differentiate the left side: The derivative of \(x\) is 1. For \(\sqrt{xy}\), use the chain rule: \(\frac{d}{dx}(\sqrt{xy}) = \frac{1}{2\sqrt{xy}} \cdot (y + x \frac{dy}{dx})\). Set the derivative of the right side, which is a constant, to 0.
Solve the resulting equation for \(\frac{dy}{dx}\) to find the slope of the tangent line at the point (4, 1). Substitute x = 4 and y = 1 into the derivative to find the specific slope at this point.
Once you have the slope of the tangent line, use the point-slope form of a line equation, \(y - y_1 = m(x - x_1)\), where m is the slope and (x_1, y_1) is the point (4, 1), to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope as the slope of the normal line, and apply the point-slope form again to find its equation.

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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 differentiate equations where the dependent and independent variables are not explicitly separated. In this case, we differentiate both sides of the equation with respect to x, treating y as a function of x. This allows us to find dy/dx, which is essential for determining the slope of the tangent line at a specific point.
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Finding The Implicit Derivative

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. To find the equation of the tangent line, we use the point-slope form of a line, which requires the slope and the coordinates of the point of tangency.
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05:13
Slopes of Tangent Lines

Normal Line

The normal line to a curve at a given point is perpendicular to the tangent line at that point. Its slope is the negative reciprocal of the slope of the tangent line. To find the equation of the normal line, we again use the point-slope form, substituting the coordinates of the point and the calculated slope of the normal line.
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Slopes of Tangent Lines
Related Practice
Textbook Question

Parallel tangent lines Find the two points where the curve x² + xy + y² = 7 crosses the x-axis, and show that the tangent lines to the curve at these points are parallel. What is the common slope of these tangent lines?

Textbook Question

Derivatives in Differential Form


In Exercises 17–28, find dy.


y = (2√x)/(3(1 + √x))

Textbook Question

In Exercises 19–22, find the slope of the curve at the point indicated.


y = x³ − 2x + 7, x = −2

Textbook Question

[Technology Exercise]


Graph the curves in Exercises 39–48.

a. Where do the graphs appear to have vertical tangent lines?

b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38.


y = 4x²/⁵ − 2x

Textbook Question

One-Sided Derivatives


Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.

Textbook Question

Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is,

d/dx (x⁻ᵐ) = −mx⁻ᵐ⁻¹

where m is a positive integer.