Textbook Question
Derivatives Find the derivative of the following functions. See Example 2 of Section 3.2 for the derivative of √x.
f(v) = v¹⁰⁰+e^v+10
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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 3.45
Derivative calculations Evaluate the derivative of the following functions at the given point.
f(s) = 2√s-1; a=25
Verified step by step guidance1
Step 1: Identify the function and the point at which you need to evaluate the derivative. The function is \( f(s) = 2\sqrt{s} - 1 \) and the point is \( a = 25 \).
Step 2: Rewrite the function in a form that is easier to differentiate. Express \( \sqrt{s} \) as \( s^{1/2} \). Thus, the function becomes \( f(s) = 2s^{1/2} - 1 \).
Step 3: Differentiate the function using the power rule. The power rule states that \( \frac{d}{ds}[s^n] = ns^{n-1} \). Apply this to \( 2s^{1/2} \) to find \( f'(s) \).
Step 4: Calculate the derivative \( f'(s) \). Using the power rule, \( f'(s) = 2 \cdot \frac{1}{2}s^{-1/2} = s^{-1/2} \).
Step 5: Evaluate the derivative at \( s = 25 \). Substitute \( s = 25 \) into \( f'(s) = s^{-1/2} \) to find \( f'(25) \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative
The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. In practical terms, the derivative at a point gives the slope of the tangent line to the function's graph at that point.
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Derivatives
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. It states that if a function is composed of two functions, say f(g(x)), the derivative can be found by multiplying the derivative of the outer function f with the derivative of the inner function g. This rule is essential when dealing with functions that involve nested expressions.
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Square Root Function
The square root function, denoted as √s, is a specific type of function that returns the non-negative value whose square is s. When differentiating functions involving square roots, it is important to apply the power rule appropriately, as √s can be rewritten as s^(1/2). Understanding how to manipulate and differentiate square root functions is crucial for solving problems involving them.
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Related Practice
Textbook Question
67–78. Derivatives of inverse functions Consider the following functions (on the given interval, if specified). Find the derivative of the inverse function.
f(x) = e^3x+1
Textbook Question
27–40. Implicit differentiation Use implicit differentiation to find dy/dx.
6x³+7y³ = 13xy
Textbook Question
Simplify the expression e^xln(x²+1).
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Textbook Question
Tangent lines Find an equation of the line tangent to the graph of f at the given point.
f(x) = sec−1(ex); (ln 2,π/3)
Textbook Question
Find the slope of the line tangent to the graph of f(x) = x / x+6 at the point (3, 1/3) and at (-2, -1/2).