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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 3.8.23b
Chapter 3, Problem 3.8.23b

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)

Verified step by step guidance
1
Start by understanding the given equation: \( \sqrt[3]{x} + \sqrt[3]{y^4} = 2 \). This is an implicit function of \( x \) and \( y \).
Differentiate both sides of the equation with respect to \( x \). Remember to use the chain rule for \( \sqrt[3]{y^4} \), which involves differentiating \( y \) with respect to \( x \) (i.e., \( \frac{dy}{dx} \)).
The derivative of \( \sqrt[3]{x} \) with respect to \( x \) is \( \frac{1}{3}x^{-2/3} \). For \( \sqrt[3]{y^4} \), apply the chain rule: \( \frac{4}{3}y^{3} \cdot \frac{dy}{dx} \).
Set the derivative of the left side equal to the derivative of the right side, which is zero, since the derivative of a constant (2) is zero.
Substitute \( x = 1 \) and \( y = 1 \) into the differentiated equation to solve for \( \frac{dy}{dx} \), which represents the slope of the curve at the point (1,1).

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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 that cannot be easily rearranged.
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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 dependent variable with respect to the independent variable at that point. Mathematically, it is found by evaluating the derivative of the function at the specified coordinates. In the context of implicit differentiation, the slope can be determined by substituting the coordinates of the point into the derivative obtained from the implicit differentiation process.
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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 encounter nested functions.
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Intro to the Chain Rule
Related Practice
Textbook Question

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)

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

1
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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}}\(\frac{\left(x^2-3\right)^5-1}{x-2}\)

Textbook Question

Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = 1/3x-1; a= 2