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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.8.20b
Chapter 3, Problem 3.8.20b

13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
tan xy = x+y; (0,0)

Verified step by step guidance
1
Start by differentiating both sides of the equation with respect to x. The equation is \( \tan(xy) = x + y \). Use implicit differentiation, which involves differentiating each term while treating y as a function of x.
Differentiate the left side: The derivative of \( \tan(u) \) with respect to u is \( \sec^2(u) \), so apply the chain rule: \( \frac{d}{dx}[\tan(xy)] = \sec^2(xy) \cdot \frac{d}{dx}(xy) \).
Apply the product rule to differentiate \( xy \): \( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \). Therefore, the derivative of the left side becomes \( \sec^2(xy) \cdot (x \frac{dy}{dx} + y) \).
Differentiate the right side: The derivative of \( x + y \) with respect to x is \( 1 + \frac{dy}{dx} \).
Set the derivatives equal to each other: \( \sec^2(xy) \cdot (x \frac{dy}{dx} + y) = 1 + \frac{dy}{dx} \). Solve this equation for \( \frac{dy}{dx} \) to find the slope of the curve at the point (0,0). Substitute x = 0 and y = 0 into the equation to find the specific slope at that point.

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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. Instead of solving for one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, applying the chain rule as necessary. This method is particularly useful for curves defined by equations like 'tan(xy) = x + y', where y cannot be easily isolated.
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Finding The Implicit Derivative

Slope of a Curve

The slope of a curve at a given point represents the rate of change of the function at that point, which is mathematically expressed as the derivative. For a curve defined implicitly, the slope can be found by evaluating the derivative obtained through implicit differentiation. In this case, we will find the slope at the point (0,0) by substituting these coordinates into the derivative expression.
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Summary of Curve Sketching

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that if a function y is defined as a function of u, which in turn is a function of x, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This rule is essential in implicit differentiation, where we often need to differentiate terms involving products of variables.
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Related Practice
Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

³√x+³√y⁴ = 2;(1,1)

Textbook Question

{Use of Tech} Angle of elevation A small plane, moving at 70 m/s, flies horizontally on a line 400 meters directly above an observer. Let θ be the angle of elevation of the plane (see figure). <IMAGE>


b. Graph dθ/dx as a function of x and determine the point at which θ changes most rapidly.

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Textbook Question

For what values of x does g(x) = x−sin x have a slope of 1?

Textbook Question

{Use of Tech} Power and energy The total energy in megawatt-hr (MWh) used by a town is given by E(t) = 400t+2400/π sin πt/12, where t≥0 is measured in hours, with t=0 corresponding to noon.

b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day?

Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(t) = 1/√t; a=9, 1/4

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Textbook Question

109-112 {Use of Tech} Calculating limits The following limits are the derivatives of a composite function g at a point a.

b. Use the Chain Rule to find each limit. Verify your answer by using a calculator.

limx→2(x2−3)5−1x−2{\(\displaystyle\]\lim\)_{x\(\to\)2}}\(\frac{\left(x^2-3\right)^5-1}{x-2}\)