Textbook Question
Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².
a. Find dr/dh for a cone with a lateral surface area of A=1500π.
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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 3.8.20a
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
tan xy = x+y; (0,0)
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. The equation is \( \tan(xy) = x + y \).
Apply the chain rule to differentiate \( \tan(xy) \). The derivative of \( \tan(u) \) with respect to u is \( \sec^2(u) \), so the derivative of \( \tan(xy) \) with respect to x is \( \sec^2(xy) \cdot (y + x \frac{dy}{dx}) \).
Differentiate the right side of the equation. The derivative of \( x \) with respect to x is 1, and the derivative of \( y \) with respect to x is \( \frac{dy}{dx} \).
Set the derivatives equal to each other: \( \sec^2(xy) \cdot (y + x \frac{dy}{dx}) = 1 + \frac{dy}{dx} \).
Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. Substitute \( x = 0 \) and \( y = 0 \) into the equation to find the specific value of \( \frac{dy}{dx} \) at the point (0,0).
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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 equations involving multiple variables or when isolating one variable is complex.
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Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. It states that if a function y is composed of another function u, 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 helps to differentiate terms involving products of variables.
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05:02Intro to the Chain Rule
Evaluating at a Point
Evaluating a derivative at a specific point involves substituting the coordinates of that point into the derivative expression. In the context of the given problem, after finding dy/dx using implicit differentiation, we substitute the point (0,0) to determine the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific values and for applications such as finding tangent lines.
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04:50Critical Points
Related Practice
Textbook Question
62–65. {Use of Tech} Graphing f and f'
a. Graph f with a graphing utility.
f(x) = (x−1) sin^−1 x on [−1,1]
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Textbook Question
A woman attached to a bungee cord jumps from a bridge that is 30 m above a river. Her height in meters above the river t seconds after the jump is y(t) = 15(1+e-t cos t), for t ≥ 0.
Determine her velocity at t = 1 and t = 3.
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Textbook Question
Comparing velocities Two stones are thrown vertically upward, each with an initial velocity of 48 ft/s at time t=0. One stone is thrown from the edge of a bridge that is 32 feet above the ground, and the other stone is thrown from ground level. The height above the ground of the stone thrown from the bridge after t seconds is f(t) = − 16t²+48t+32. and the height of the stone thrown from the ground after t seconds is g(t) = −16t²+48t.
a. Show that the stones reach their high points at the same time.
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Textbook Question
Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.
f(x) = 1/x; P (1,1)
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Textbook Question
Vertical tangent lines If a function f is continuous at a and lim x→a| f′(x)|=∞, then the curve y=f(x) has a vertical tangent line at a, and the equation of the tangent line is x=a. If a is an endpoint of a domain, then the appropriate one-sided derivative (Exercises 71–72) is used. Use this information to answer the following questions.
73. {Use of Tech} Graph the following functions and determine the location of the vertical tangent lines.
a. f(x) = (x-2)^1/3
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