Textbook Question
79–82. {Use of Tech} Visualizing tangent and normal lines
b. Graph the tangent and normal lines on the given graph.
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)
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 3.8.14b
13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
x = e^y; (2, ln 2)
Verified step by step guidance1
Start by recognizing that the equation given is in implicit form: \( x = e^y \). This means that \( y \) is not isolated on one side of the equation.
To find the derivative \( \frac{dy}{dx} \), apply implicit differentiation to both sides of the equation with respect to \( x \). Differentiate \( x \) to get 1, and differentiate \( e^y \) using the chain rule to get \( e^y \cdot \frac{dy}{dx} \).
Set up the equation from the differentiation: \( 1 = e^y \cdot \frac{dy}{dx} \).
Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation: \( \frac{dy}{dx} = \frac{1}{e^y} \).
Substitute the given point \((2, \ln 2)\) into the expression for \( \frac{dy}{dx} \). Since \( y = \ln 2 \), calculate \( e^{\ln 2} \) which simplifies to 2, and find the slope at this 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 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 function at that point, which is mathematically defined as the derivative of the function. For a curve defined implicitly, the slope can be found by evaluating the derivative obtained through implicit differentiation. The slope is often denoted as 'dy/dx' and indicates how steep the curve is at the specified coordinates.
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11:41Summary of Curve Sketching
Exponential and Natural Logarithm Functions
Exponential functions, such as e^y, and natural logarithm functions, like ln(x), are fundamental in calculus. The function e^y is the inverse of ln(y), and they exhibit unique properties, such as the derivative of e^y being e^y dy/dx. Understanding these functions is crucial for evaluating expressions and derivatives involving exponential growth or decay, especially when working with implicit relationships.
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05:18Derivative of the Natural Logarithmic Function
Related Practice
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.
b. Evaluate this derivative when r=30 and h=40.
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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) = √3x; a= 12
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Textbook Question
13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
cos y = x; (0, π/2)
Textbook Question
{Use of Tech} Spring oscillations A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance x (in inches) of the mass from its equilibrium position after t seconds is given by the function x(t) = 10sin t - 10cos t, where x is positive when the mass is above the equilibrium position. <IMAGE>
b. Find dx/dt and interpret the meaning of this derivative.
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Textbook Question
21–30. Derivatives
b. Evaluate f'(a) for the given values of a.
f(x) = 1/x+1; a = -1/2;5