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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.5.87.c

Derivatives of sin^n x Calculate the following derivatives using the Product Rule.
c. d/dx (sin⁴ x)

Verified step by step guidance
1
First, recognize that the function sin⁴(x) can be rewritten as (sin(x))⁴. This helps in applying the chain rule effectively.
Apply the chain rule: If you have a function u(x) raised to a power n, the derivative is n * u(x)^(n-1) * u'(x). Here, u(x) = sin(x) and n = 4.
Calculate the derivative of u(x) = sin(x), which is u'(x) = cos(x).
Substitute u(x) = sin(x) and u'(x) = cos(x) into the chain rule formula: 4 * (sin(x))³ * cos(x).
Simplify the expression to get the derivative: 4 * sin³(x) * cos(x). This is the derivative of sin⁴(x) using the chain rule.

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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 technique used when finding the derivative of a product of two functions. It states that if you have two functions, u(x) and v(x), the derivative of their product is given by d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). This rule is essential for calculating derivatives where functions are multiplied together, such as in the case of sin^n x.
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The Product Rule

Chain Rule

The Chain Rule is another critical differentiation rule used when dealing with composite functions. It states that if a function y is composed of another function u, such that y = f(u) and u = g(x), then the derivative is given by dy/dx = dy/du * du/dx. In the context of sin^n x, the Chain Rule is necessary to differentiate the inner function (sin x) raised to a power.
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Intro to the Chain Rule

Higher Order Derivatives

Higher order derivatives refer to the derivatives of a function taken multiple times. For example, the second derivative is the derivative of the first derivative. In the context of sin^n x, understanding higher order derivatives can be important for analyzing the behavior of the function, such as concavity and points of inflection, especially when applying the Product Rule and Chain Rule in more complex scenarios.
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Higher Order Derivatives
Related Practice
Textbook Question

{Use of Tech} Power and energy 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 / 9kWh where t=0 corresponds to midnight.

c. Graph the power function and interpret the graph. What are the units of power in this case?

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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>

c. (f^-1)'(1)

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Textbook Question

62–65. {Use of Tech} Graphing f and f'

c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.

f(x)=(sec^−1 x)/x on [1,∞)

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Textbook Question

Derivatives from a graph If possible, evaluate the following derivatives using the graphs of f and f'. <IMAGE>

c. (f^-1)'(f(2))

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Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t²+19.6t+24.5.

c. What is the height of the stone at the highest point?

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Textbook Question

Vibrations of a spring Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position y=0y=0 when the mass hangs at rest. Suppose you push the mass to a position y0y_0 units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is y=y0cos(tkm)y=y_0\(\cos\]\left\)(t\(\sqrt{\frac{k}{m}\)}\(\right\)), where k>0k>0 is a constant measuring the stiffness of the spring (the larger the value of kk, the stiffer the spring) and yy is positive in the upward direction.

Use equation (4) to answer the following questions.

c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness (kk is increased by a factor of 44)?

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