Textbook Question
Calculate the derivative of the following functions.
y = (1 - e0.05x)-1
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 57
Find and simplify the derivative of the following functions.
h(x) = (5x7 + 5x)(6x3 + 3x2 + 3)
Verified step by step guidance1
Step 1: Identify the function h(x) = (5x^7 + 5x)(6x^3 + 3x^2 + 3) as a product of two functions, u(x) = 5x^7 + 5x and v(x) = 6x^3 + 3x^2 + 3.
Step 2: Apply the product rule for derivatives, which states that if h(x) = u(x)v(x), then h'(x) = u'(x)v(x) + u(x)v'(x).
Step 3: Differentiate u(x) = 5x^7 + 5x to find u'(x). Use the power rule: d/dx[x^n] = nx^(n-1).
Step 4: Differentiate v(x) = 6x^3 + 3x^2 + 3 to find v'(x). Again, use the power rule for each term.
Step 5: Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula to find h'(x) and simplify the expression.
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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 calculus, the derivative is often denoted as f'(x) or df/dx, and it provides critical information about the function's behavior, such as its slope and points of tangency.
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Product Rule
The product rule is a formula used to find the derivative of the product of two functions. If u(x) and v(x) are two differentiable functions, the product rule states that the derivative of their product is given by u'v + uv'. This rule is essential when differentiating functions that are expressed as products, allowing for the correct application of the derivative to each component.
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Simplification of Derivatives
After finding the derivative of a function, simplification is often necessary to express the result in a more manageable form. This may involve combining like terms, factoring, or reducing fractions. Simplifying the derivative can help in analyzing the function's behavior, such as identifying critical points and understanding the function's increasing or decreasing intervals.
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Related Practice
Textbook Question
Determining the unknown constant Let f(x) = {2x² if x≤1 ax-2 if x>1. Determine a value of a (if possible) for which f' is continuous at x=1.
Textbook Question
The energy (in joules) released by an earthquake of magnitude M is given by the equation E = 25,000 ⋅ 101.5M. (This equation can be solved for M to define the magnitude of a given earthquake; it is a refinement of the original Richter scale created by Charles Richter in 1935.)
Compute the energy released by earthquakes of magnitude 1, 2, 3, 4, and 5. Plot the points on a graph and join them with a smooth curve.
Textbook Question
{Use of Tech} Equations of tangent lines
Find an equation of the line tangent to the given curve at a.
y = −3x2 + 2; a=1
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Textbook Question
Power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up. It is measured in units of joules or Calories, where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶J, or 240 Cal. On the other hand, power is the rate at which energy is used, which is measured in watts, where 1 W = 1 J/s. Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for one hour, the total amount of energy used is 1 kilowatt-hour (1 kWh = 3.6×10⁶ J) Suppose the cumulative energy used in a large building over a 24-hr period is given by E(t)=100t + 4t² − (t³ / 9) kWh where t = 0 corresponds to midnight.
The power is the rate of energy consumption; that is, P(t) = E′(t) Find the power over the interval 0 ≤ t ≤ 24.
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Textbook Question
Calculate the derivative of the following functions.
y = cos7/4(4x3)