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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.

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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:
  • \( h'(x) = f'(g(x)) \cdot g'(x) \)
In this problem, for example, consider \( h(s) = \sin^3 s \). Here, \( f(u) = u^3 \) and \( g(s) = \sin s \). To find the derivative using the chain rule, we compute:
  • 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.
    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 \( h(s) = \sin^3 s + \cos^3 s \):
    • \( \sin(s) \) 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.
    In our specific problem, these trigonometric functions \( \sin^3 s \) and \( \cos^3 s \) are raised to a power, which means we must first consider the power rule before applying the essential chain rule to get the derivatives.
    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:
    • 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.
    In the given function \( h(s) = \sin^3 s + \cos^3 s \), finding the derivative requires us to understand the change in both components: \( \sin^3 s \) and \( \cos^3 s \).

    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 \( h(s) \).
    This approach allows us to understand and predict the behavior of complex functions more effectively.

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    Most popular questions from this chapter

    Use the formal definition to find the derivative of \(y=\frac{1}{x}\) at \(x=2\). (b) Show that the point \(\left(2, \frac{1}{2}\right)\) is on the graph of \(y=\frac{1}{x}\), and find the equation of the normal line at the point \(\left(2, \frac{1}{2}\right)\). (c) Graph \(y=\frac{1}{x}\) and the tangent line at the point \(\left(2, \frac{1}{2}\right)\) in the same coordinate system.

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    Graph each function and, on the basis of the graph, guess where the function is not differentiable. (Assume the largest possible domain.) $$ f(x)=\left\\{\begin{array}{cl} x & \text { for } x \leq 0 \\ x+1 & \text { for } x>0 \end{array}\right. $$

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