Textbook Question
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
cos y = x; (0, π/2)
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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 3.39a
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = √2x+1; a= 4
Verified step by step guidance1
Step 1: Identify the function f(x) = \(\sqrt{2x + 1}\) and the point a = 4 where we need to find the derivative f'(a).
Step 2: Use the chain rule to differentiate f(x). The chain rule states that if you have a composite function f(g(x)), then the derivative is f'(g(x)) * g'(x).
Step 3: Recognize that f(x) = (2x + 1)^{1/2}. Differentiate the outer function (2x + 1)^{1/2} with respect to (2x + 1), which gives (1/2)(2x + 1)^{-1/2}.
Step 4: Differentiate the inner function 2x + 1 with respect to x, which gives 2.
Step 5: Combine the results from Steps 3 and 4 using the chain rule: f'(x) = (1/2)(2x + 1)^{-1/2} * 2. Simplify this expression and then substitute x = 4 to find f'(4).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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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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 around specific values.
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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 f(a) and then apply the derivative rules.
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4:26Evaluating Composed Functions
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
21–30. Derivatives
a. Use limits to find the derivative function f' for the following functions f.
f(x) = 1/x+1; a = -1/2;5
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Textbook Question
Tracking a dive A biologist standing at the bottom of an 80-foot vertical cliff watches a peregrine falcon dive from the top of the cliff at a 45° angle from the horizontal (see figure). <IMAGE>
a. Express the angle of elevation θ from the biologist to the falcon as a function of the height h of the bird above the ground. (Hint: The vertical distance between the top of the cliff and the falcon is 80−h.)
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Textbook Question
21–30. Derivatives
a. Use limits to find the derivative function f' for the following functions f.
f(t) = 3t⁴; a= -2, 2
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