Textbook Question
9–61. Evaluate and simplify y'.
y = e^tan x (tan x−1)
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.R.70
66–71. Higher-order derivatives Find and simplify y''.
x + sin y = y
Verified step by step guidance1
First, identify the given equation: \( x + \sin(y) = y \). We need to find the second derivative \( y'' \).
Differentiate both sides of the equation with respect to \( x \) to find the first derivative \( y' \). Remember to use the chain rule for \( \sin(y) \), which gives \( \cos(y) \cdot y' \).
The differentiation of the left side \( x + \sin(y) \) with respect to \( x \) results in \( 1 + \cos(y) \cdot y' \). The right side \( y \) differentiates to \( y' \). Set these equal: \( 1 + \cos(y) \cdot y' = y' \).
Solve for \( y' \) by isolating it on one side of the equation. This involves rearranging the terms to get \( 1 = y' - \cos(y) \cdot y' \), which simplifies to \( 1 = y'(1 - \cos(y)) \).
Now, differentiate \( y' = \frac{1}{1 - \cos(y)} \) with respect to \( x \) to find \( y'' \). Use the quotient rule and chain rule as needed, and simplify the expression to obtain \( y'' \).
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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 have an equation involving both x and y, and we need to differentiate with respect to x while treating y as a function of x. This involves applying the chain rule to account for the derivatives of y.
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Finding The Implicit Derivative
Higher-Order Derivatives
Higher-order derivatives refer to the derivatives of a function beyond the first derivative. The second derivative, denoted as y'', provides information about the curvature of the function and can indicate concavity. To find y'', we first need to find the first derivative y' and then differentiate it again, applying implicit differentiation as necessary.
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02:42Higher Order Derivatives
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. When differentiating y with respect to x, we apply the chain rule to account for the fact that y is a function of x. This means that when we differentiate terms involving y, we must multiply by dy/dx, which represents the derivative of y with respect to x.
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05:02Intro to the Chain Rule
Related Practice
Textbook Question
Use the given graphs of f and g to find each derivative. <IMAGE>
b. d/dx (f(x)g(x)) |x=1
Textbook Question
Evaluate and simplify y'.
y = ln |sec 3x|
1
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Textbook Question
9–61. Evaluate and simplify y'.
y = tan (sin θ)
Textbook Question
{Use of Tech} Beak length The length of the culmen (the upper ridge of a bird’s bill) of a t-week-old Indian spotted owlet is modeled by the function L(t)=11.94 / 1 + 4e^−1.65t, where L is measured in millimeters.
b. Use a graph of L′(t) to describe how the culmen grows over the first 5 weeks of life.
Textbook Question
Use the given graphs of f and g to find each derivative. <IMAGE>
c. d/dx ((f(x) / g(x)) |x=3