Domanda del libro di testo
Vertical tangent lines
a. Determine the points where the curve x+y³−y=1 has a vertical tangent line (see Exercise 60).
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- 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: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.37a
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = 1/ √x; a= 1/4
Guida verificata passo dopo passo1
Step 1: Identify the function and the point of interest. The function given is \( f(x) = \frac{1}{\sqrt{x}} \) and we need to find the derivative at \( a = \frac{1}{4} \).
Step 2: Rewrite the function in a form that is easier to differentiate. Express \( f(x) = \frac{1}{\sqrt{x}} \) as \( f(x) = x^{-1/2} \).
Step 3: Use the power rule for differentiation. The power rule states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). Apply this to \( f(x) = x^{-1/2} \) to find \( f'(x) \).
Step 4: Differentiate \( f(x) = x^{-1/2} \) using the power rule. This gives \( f'(x) = -\frac{1}{2}x^{-3/2} \).
Step 5: Evaluate the derivative at \( x = \frac{1}{4} \). Substitute \( x = \frac{1}{4} \) into \( f'(x) = -\frac{1}{2}x^{-3/2} \) to find \( f'(\frac{1}{4}) \).
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Derivatives
A derivative represents the rate of change of a function with respect to its variable. It is defined as the limit of the average rate of change of the function as the interval approaches zero. In practical terms, the derivative at a point gives the slope of the tangent line to the function at that point.
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Percorso guidato
05:44
Derivatives
Tangent Lines
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. This concept is crucial for understanding how functions behave locally and is often used in optimization problems.
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05:13Slopes of Tangent Lines
Function Evaluation
Function evaluation involves substituting a specific value into a function to determine its output. In the context of derivatives, evaluating the function at a point helps in calculating the derivative at that point. For example, to find f′(a), one must first evaluate the function f(x) at the point a before applying derivative rules.
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4:26Evaluating Composed Functions
Pratica correlata
Domanda del libro di testo
Highway travel A state patrol station is located on a straight north-south freeway. A patrol car leaves the station at 9:00 A.M. heading north with position function s = f(t) that gives its location in miles t hours after 9:00 A.M. (see figure). Assume s is positive when the car is north of the patrol station. <IMAGE>
a. Determine the average velocity of the car during the first 45 minutes of the trip.
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Domanda del libro di testo
62–65. {Use of Tech} Graphing f and f'
a. Graph f with a graphing utility.
f(x)=e^−x tan^−1 x on [0,∞)
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Domanda del libro di testo
Derivatives of inverse functions from a table Use the following tables to determine the indicated derivatives or state that the derivative cannot be determined. <IMAGE>
a. (f^-1)'(4)
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Domanda del libro di testo
Use the definition of the derivative to determine d/dx(ax²+bx+c), where a, b, and c are constants.
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Domanda del libro di testo
79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)
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