Textbook Question
Find an equation of the line tangent to the following curves at the given value of x.
y = 4 sin x cos x; x = π/3
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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 3.8.17a
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
sin y = 5x⁴−5; (1, π)
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. The equation is \( \sin y = 5x^4 - 5 \).
For the left side, differentiate \( \sin y \) with respect to y, which gives \( \cos y \), and then multiply by \( \frac{dy}{dx} \) due to the chain rule. This results in \( \cos y \cdot \frac{dy}{dx} \).
For the right side, differentiate \( 5x^4 - 5 \) with respect to x. The derivative of \( 5x^4 \) is \( 20x^3 \), and the derivative of \(-5\) is 0. So, the right side becomes \( 20x^3 \).
Set the derivatives equal to each other: \( \cos y \cdot \frac{dy}{dx} = 20x^3 \).
Solve for \( \frac{dy}{dx} \) by dividing both sides by \( \cos y \), resulting in \( \frac{dy}{dx} = \frac{20x^3}{\cos y} \).
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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. Instead of solving for y in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. This method is particularly useful for equations that are difficult or impossible to rearrange.
Recommended video:
05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When using implicit differentiation, the chain rule is applied to terms involving y, resulting in the derivative dy/dx. This means that when differentiating a function of y, we multiply by dy/dx to account for the dependence of y on x.
Recommended video:
05:02Intro to the Chain Rule
Evaluating at a Point
After finding the derivative dy/dx using implicit differentiation, we often need to evaluate it at a specific point, such as (1, π) in this case. This involves substituting the x and y values into the derived expression to find the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific coordinates.
Recommended video:
04:50Critical Points
Related Practice
Textbook Question
45–50. Tangent lines Carry out the following steps. <IMAGE>
a. Verify that the given point lies on the curve.
x³+y³=2xy; (1, 1)
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Textbook Question
7–14. Find the derivative the following ways:
a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.
f(w) = w³ -w / w
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Textbook Question
Airline travel The following figure shows the position function of an airliner on an out-and-back trip from Seattle to Minneapolis, where s = f(t) is the number of ground miles from Seattle t hours after take-off at 6:00 A.M. The plane returns to Seattle 8.5 hours later at 2:30 P.M. <IMAGE>
a. Calculate the average velocity of the airliner during the first 1.5 hours of the trip (0 ≤ t ≤ 1.5).
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Textbook Question
Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of 4 m/s.
a. At what rate is the area of the triangle changing when the legs are 5 m long?
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Textbook Question
{Use of Tech} Approximating derivatives Assuming the limit exists, the definition of the derivative f′(a) = lim h→0 f(a + h) − f(a) / h implies that if ℎ is small, then an approximation to f′(a) is given by
f' (a) ≈ f(a+h) - f(a) / h. If ℎ > 0 , then this approximation is called a forward difference quotient; if ℎ < 0 , it is a backward difference quotient. As shown in the following exercises, these formulas are used to approximate f′ at a point when f is a complicated function or when f is represented by a set of data points. <IMAGE>
Let f (x) = √x.
a. Find the exact value of f' (4).
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