Chapter 4: Problem 51
In Problems \(1-58\), find the derivative with respect to the independent variable. $$ h(s)=\sin ^{3} s+\cos ^{3} s $$
Short Answer
Expert verified
The derivative is \( 3\sin s \cdot \cos s (\sin s - \cos s) \).
Step by step solution
01
Understand the Function
The function given is \( h(s) = \sin^3 s + \cos^3 s \). This means we have two terms: \( \sin^3 s \) and \( \cos^3 s \). Each term is a trigonometric function raised to the power of 3.
02
Apply the Chain Rule
To find the derivative of \( \sin^3 s \), we use the chain rule. The chain rule states that if you have a composition of functions \( f(g(x)) \), then the derivative is \( f'(g(x)) \cdot g'(x) \). Here, \( f(u) = u^3 \) and \( g(s) = \sin s \), so its derivative is \( 3\sin^2 s \cdot \cos s \).
03
Differentiate the Cosine Term
Similarly, for \( \cos^3 s \), we first find the derivative using the chain rule. Let \( f(u) = u^3 \) and \( g(s) = \cos s \). The derivative is \( 3\cos^2 s \cdot (-\sin s) \), which simplifies to \(-3\cos^2 s \cdot \sin s \).
04
Combine the Derivatives
Now, combine the derivatives of both parts. The derivative of \( h(s) \) is \( 3\sin^2 s \cdot \cos s - 3\cos^2 s \cdot \sin s \).
05
Factor the Expression
Factor out the common terms in the expression: \( 3\sin s \cdot \cos s \left( \sin s - \cos s \right) \). This is the simplified form of the derivative.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule in Calculus
The chain rule is a fundamental technique in calculus used to differentiate the composition of two or more functions. Imagine you have a function made up of a function inside another function, like a matryoshka doll within another. If you want to differentiate this nested function form, the chain rule helps with this task.
Suppose you have a composite function, say, \( h(x) = f(g(x)) \). The chain rule formula tells us that the derivative of \( h(x) \) is the derivative of \( f \) evaluated at \( g(x) \), multiplied by the derivative of \( g(x) \). Mathematically, this is expressed as:Differentiate \( f(u) = u^3 \) to get \( f'(u) = 3u^2 \); this means \( f'\left( \sin(s) \right) = 3(\sin s)^2 \). Differentiate \( g(s) = \sin s \) to get \( g'(s) = \cos s \). Combine these to get the derivative for the term \( 3\sin^2 s \cdot \cos s \). The chain rule makes handling complex compositions manageable by breaking them down into simpler parts.
Suppose you have a composite function, say, \( h(x) = f(g(x)) \). The chain rule formula tells us that the derivative of \( h(x) \) is the derivative of \( f \) evaluated at \( g(x) \), multiplied by the derivative of \( g(x) \). Mathematically, this is expressed as:
- \( h'(x) = f'(g(x)) \cdot g'(x) \)
Understanding Trigonometric Functions
Trigonometric functions like sine and cosine are foundational in calculus and various fields of mathematics. They relate the angles of a triangle to the lengths of its sides and have applications extending far beyond geometry. These functions are periodic, repetitive over intervals, typically with a periodicity of \( 2\pi \).
Let's break down the terms given in the function \( h(s) = \sin^3 s + \cos^3 s \):
Let's break down the terms given in the function \( h(s) = \sin^3 s + \cos^3 s \):
- \( \sin(s) \) and \( \cos(s) \) represent the vertical and horizontal components, respectively, in the unit circle, which are crucial in defining these functions.
- Both \( \sin(s) \) and \( \cos(s) \) oscillate between -1 and 1.
- Their derivatives are \( \cos(s) \) and \(-\sin(s) \), respectively. These derivatives are pivotal in calculus as they describe the rate of change of the sine and cosine functions.
Derivatives and Their Importance
Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. Simply put, it is the slope of a function at a particular point on its graph. The derivative helps describe how one quantity changes in relation to another.
In practical terms, derivatives can be understood as:
Using the chain rule and knowing the basic derivatives of \( \sin(s) \) and \( \cos(s) \), we get:
In practical terms, derivatives can be understood as:
- Indicating the steepness or incline of a function on a graph.
- Helping to understand acceleration as the derivative of velocity concerning time.
- Finding maximum and minimum values of functions, essential in optimization problems.
Using the chain rule and knowing the basic derivatives of \( \sin(s) \) and \( \cos(s) \), we get:
- Derivative of \( \sin^3 s \) is \( 3\sin^2 s \cdot \cos s \).
- Derivative of \( \cos^3 s \) is \(-3\cos^2 s \cdot \sin s \).
- Combining these using simple algebraic steps, we can find the derivative of the entire expression \( h(s) \).