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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 3.8.13b
Chapter 3, Problem 3.8.13b

13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.x⁴+y⁴ = 2;(1,−1)

Verified step by step guidance
1
Start by differentiating both sides of the equation with respect to x. The equation is x⁴ + y⁴ = 2. Use implicit differentiation, which involves differentiating y with respect to x as well.
Differentiate x⁴ with respect to x, which gives 4x³. For y⁴, apply the chain rule: differentiate y⁴ with respect to y to get 4y³, then multiply by dy/dx (the derivative of y with respect to x).
Set up the equation from the differentiation: 4x³ + 4y³(dy/dx) = 0. This equation represents the derivative of the original equation.
Solve for dy/dx, which represents the slope of the curve. Rearrange the equation to isolate dy/dx: dy/dx = -4x³ / 4y³.
Substitute the given point (1, -1) into the equation for dy/dx. Replace x with 1 and y with -1 to find the slope at this specific 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 x⁴ + y⁴ = 2, 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 y-coordinate with respect to the x-coordinate at that point. Mathematically, it is found by evaluating the derivative of the function at the specified point. In the context of implicit differentiation, once we find dy/dx, we can substitute the coordinates of the point (1, -1) to determine the slope of the curve at that location.
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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, as it allows us to differentiate terms involving y when applying the derivative to both sides of an equation.
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Related Practice
Textbook Question

62–65. {Use of Tech} Graphing f and f'

b. Compute and graph f'.

f(x) = (x−1) sin^−1 x on [−1,1]

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.

limh→013((1+h)5+7)10−13(8)10h{\(\displaystyle\]\lim\)_{h\(\to\)0}}\(\frac{\frac{1}{3\left(\left(1+h\right)^5+7\right)^{10}\)}-\(\frac{1}{3\left(8\right)^{10}\)}}{h}

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Textbook Question

21–30. Derivatives

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

f(s) = 4s³+3s; a= -3, -1

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Textbook Question

Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+32t+48.

b. When does the stone reach its highest point?

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Textbook Question

Product Rule for three functions Assume f, g, and h are differentiable at x.

b. Use the formula in (a) to find d/dx(e^x(x−1)(x+3))

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Textbook Question

Tracking a dive A biologist standing at the bottom of an 80-foot vertical cliff watches a peregrine falcon dive from the top of the cliff at a 45° angle from the horizontal (see figure). <IMAGE>


b. What is the rate of change of θ with respect to the bird’s height when it is 60 ft above the ground?

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