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Chapter 3: Problem 29

Find the derivatives of the functions. Assume \(a, b,\) and \(c\) are constants.$$y=\sin ^{5} \theta$$

Short Answer

Expert verified
The derivative is \( \frac{dy}{d\theta} = 5 \sin^4 \theta \cdot \cos \theta \).

Step by step solution

01

Identify the Function

The function given is \[y = \sin^5 \theta\]This can be rewritten as \[y = (\sin \theta)^5\]This is a composite function, which suggests using the chain rule for differentiation.
02

Apply the Chain Rule

The chain rule states that if you have a composite function \(y = f(g(x))\), then the derivative is \[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]Here, \(y = (\sin \theta)^5\), where \(u = \sin \theta\) and \(f(u) = u^5\). We'll have to find the derivative of \(u\) first, which is \(\cos \theta\), and the derivative of \(u^5\), which is \(5u^4\).
03

Differentiate the Outer Function

Differentiate \(u^5\) with respect to \(u\):\[\frac{d}{du}(u^5) = 5u^4\]Set it as your first part.
04

Differentiate the Inner Function

Differentiate \(\sin \theta\) with respect to \(\theta\):\[\frac{d}{d\theta}(\sin \theta) = \cos \theta\]This becomes your second part.
05

Combine Derivatives Using the Chain Rule

Multiply the derivatives of the outer and inner functions as per the chain rule:\[ \frac{dy}{d\theta} = 5(\sin \theta)^4 \cdot \cos \theta \]
06

Simplify the Expression

Thus, the derivative of the function simplifies to:\[ \frac{dy}{d\theta} = 5 \sin^4 \theta \cdot \cos \theta \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
When dealing with composite functions, the chain rule is an essential tool for differentiation. In simple terms, the chain rule allows us to differentiate a function that is composed of other functions. It helps us break down complicated expressions into manageable parts.

Suppose you have a composite function like \(y = f(g(x))\). The chain rule states that the derivative of this function can be found using the formula:
  • \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)
This rule breaks the process into two main steps:
- **Differentiate the outer function** while treating the inner function as constant.- **Differentiate the inner function** with respect to the variable.

Once you have these two derivatives, multiply them to find the derivative of the composite function. This concept is crucial when differentiating more complicated expressions, such as those involving trigonometric functions raised to a power.
Composite Function
A composite function is a combination of two or more functions. It takes the output of one function and uses it as the input for another. In mathematical terms, if you have functions \(f(x)\) and \(g(x)\), their composition is denoted as \(f(g(x))\).

This means that you first apply the function \(g(x)\) and then apply \(f(x)\) to the result. For example, in the function \(y = \sin^5 \theta\), it can be broken down into:
  • **Inner Function**: \(u = \sin \theta\)
  • **Outer Function**: \(f(u) = u^5\)
By treating \(\sin \theta\) as the inner function and \(u^5\) as the outer function, you can apply the chain rule to find its derivative. This approach simplifies the differentiation process for complex functions, allowing you to tackle each part separately.
Trigonometric Differentiation
Trigonometric functions like sine, cosine, and tangent are foundational in calculus, and their derivatives are essential to understand. Derivatives of basic trigonometric functions are part of fundamental calculus rules.
  • The derivative of \(\sin \theta\) is \(\cos \theta\).
  • The derivative of \(\cos \theta\) is \(-\sin \theta\).
  • The derivative of \(\tan \theta\) is \(\sec^2 \theta\).
In the given exercise, \(y = \sin^5 \theta\), you first see \(\sin \theta\) as the inner function. Differentiating \(\sin \theta\) yields \(\cos \theta\), which is a critical step in applying the chain rule. Knowing these trigonometric differentiation rules is necessary for solving problems involving trigonometric expressions, making it easier to understand and tackle more complex calculus tasks effectively.

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Most popular questions from this chapter

Give an example of:A trigonometric function whose derivative must be calculated using the chain rule.

Find the derivative of the function. $$f(t)=\cosh ^{2} t-\sinh ^{2} t$$

Suppose that \(f^{\prime \prime}\) and \(g^{\prime \prime}\) exist and that \(f\) and \(g\) are concave up for all \(x .\) Are the statements in Problems \(91-92\) true or false for all such \(f\) and \(g\) ? If a statement is true, explain how you know. If a statement is false, give a counterexample. \(f(x)+g(x)\) is concave up for all \(x\)

Consider the graph of \(f(x)=x^{2}\) near \(x=1 .\) Find an interval around \(x=1\) with the property that throughout any smaller interval, the graph of \(f(x)=x^{2}\) never differs from its local linearization at \(x=1\) by more than \(0.1|x-1|\)

A wave travels along a string that is joined to a thicker rope. The wave both reflects back along the string and is transmitted to the rope. For positive constants \(k_{1}, k_{2}, w\) and time \(t,\) the wave along the string, given by \(x<0,\) is $$f(x)=\cos \left(k_{1} x-w t\right)+R \cos \left(-k_{1} x-w t\right)$$and the wave along the rope, given by \(x>0,\) is $$ g(x)=T \cos \left(k_{2} x-w t\right)$$ For every value of \(t,\) the two waves have the same tangent line at \(x=0,\) so they have the same value and same slope at \(x=0 .\) Use this fact to show that $$R=\frac{k_{1}-k_{2}}{k_{1}+k_{2}} \quad \text { and } \quad T=\frac{2 k_{1}}{k_{1}+k_{2}}$$ These amplitudes are called the reflection coefficient, \(R\) and the transmission coefficient, \(T\).
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