Textbook Question
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>
b. (f^-1)'(6)
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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 3.4.99b
Product Rule for three functions Assume f, g, and h are differentiable at x.
b. Use the formula in (a) to find d/dx(e^x(x−1)(x+3))
Verified step by step guidance1
Step 1: Identify the functions involved in the product. Here, we have three functions: \(f(x) = e^x\), \(g(x) = (x - 1)\), and \(h(x) = (x + 3)\). We need to differentiate the product \(f(x)g(x)h(x)\).
Step 2: Recall the product rule for three functions. If \(u(x)\), \(v(x)\), and \(w(x)\) are differentiable functions, then the derivative of their product is given by: \(\frac{d}{dx}[u(x)v(x)w(x)] = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x)\).
Step 3: Differentiate each function individually. Compute \(f'(x)\), \(g'(x)\), and \(h'(x)\). For \(f(x) = e^x\), \(f'(x) = e^x\). For \(g(x) = (x - 1)\), \(g'(x) = 1\). For \(h(x) = (x + 3)\), \(h'(x) = 1\).
Step 4: Apply the product rule for three functions. Substitute \(f(x)\), \(g(x)\), \(h(x)\), and their derivatives into the formula: \(\frac{d}{dx}[e^x(x-1)(x+3)] = e^x \cdot (x-1)(x+3) + e^x \cdot 1 \cdot (x+3) + e^x \cdot (x-1) \cdot 1\).
Step 5: Simplify the expression. Combine like terms and simplify the expression obtained in Step 4 to get the final derivative.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Product Rule
The Product Rule is a fundamental differentiation rule used to find the derivative of the product of two or more functions. For two functions f(x) and g(x), the rule states that the derivative of their product is given by f'(x)g(x) + f(x)g'(x). This concept extends to three functions, where the derivative of f(x)g(x)h(x) is f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x).
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The Product Rule
Chain Rule
The Chain Rule is another essential differentiation technique that allows us to differentiate composite functions. If a function y is defined as a composition of two functions, such as y = f(g(x)), the Chain Rule states that the derivative is dy/dx = f'(g(x)) * g'(x). While not directly applied in the product rule, understanding the Chain Rule is crucial when dealing with functions that involve exponentials or other nested functions.
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05:02Intro to the Chain Rule
Exponential Functions
Exponential functions, such as e^x, are functions where the variable appears in the exponent. The derivative of e^x is unique because it is equal to e^x itself, making it particularly straightforward to differentiate. In the context of the given problem, recognizing that e^x is part of the product allows for easier application of the Product Rule, as its derivative does not change the form of the function.
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Related Practice
Textbook Question
{Use of Tech} Bungee jumper 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.
b. Use a graphing utility to determine when she is moving downward and when she is moving upward during the first 10 s.
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Textbook Question
31–32. Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) above the ground after t seconds is given by the function s(t).
b. Determine the instantaneous velocity of the projectile at t=1 and t = 2 seconds.
s(t)= −16t²+100t
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Textbook Question
The Chain Rule for second derivatives
b. Use the formula in part (a) to calculate .
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Textbook Question
Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+32t+48.
b. When does the stone reach its highest point?
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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>
b. What is the rate of change of θ with respect to the bird’s height when it is 60 ft above the ground?
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