Problem 5 By applying the chain rule: (a... [FREE SOLUTION]

Chapter 16: Problem 5

By applying the chain rule: (a) Write out the derivative formulas for \(\frac{d}{d \theta} \sin f(\theta)\) and \(\frac{d}{d \theta} \cos f(\theta),\) where \(f(\theta)\) is a function of \(\theta\) (b) Find the derivatives of \(\cos \theta^{3}, \sin \left(\theta^{2}+3 \theta\right), \cos e^{\prime \prime},\) and \(\sin (1 / \theta)\).

Short Answer

Expert verified

Apply the chain rule to find derivatives: \([-3\theta^2 \sin \theta^3, \cos(\theta^2 + 3\theta)(2\theta + 3), -e^{\theta} \sin e^{\theta}, -\frac{\cos(1/\theta)}{\theta^2}]\).

Step by step solution

Review the Chain Rule

The chain rule is a formula to compute the derivative of a composite function. If you have a function \( y = g(f(x)) \), where both \( f \) and \( g \) are functions of \( x \), then the derivative can be calculated as:\[ \frac{dy}{dx} = g'(f(x)) \cdot f'(x) \]

Derive Formula for \(\frac{d}{d\theta} \sin f(\theta)\)

To find the derivative of \( \sin f(\theta) \), identify \( g(u) = \sin u \) and \( u = f(\theta) \). Using the chain rule:\[ \frac{d}{d\theta} \sin f(\theta) = \cos f(\theta) \cdot f'(\theta) \]

Derive Formula for \(\frac{d}{d\theta} \cos f(\theta)\)

For \( \cos f(\theta) \), identify \( g(u) = \cos u \) and \( u = f(\theta) \). The chain rule gives us:\[ \frac{d}{d\theta} \cos f(\theta) = -\sin f(\theta) \cdot f'(\theta) \]

Differentiate \(\cos \theta^{3}\)

For \( \cos \theta^{3} \), let \( g(u) = \cos u \) and \( u = \theta^3 \). Then:\[ \frac{d}{d\theta} \cos \theta^3 = -\sin \theta^3 \cdot 3\theta^2 \]

Differentiate \(\sin(\theta^{2} + 3\theta)\)

For \( \sin(\theta^{2} + 3\theta) \), let \( g(u) = \sin u \) and \( u = \theta^{2} + 3\theta \). Then:\[ \frac{d}{d\theta} \sin(\theta^{2} + 3\theta) = \cos(\theta^{2} + 3\theta) \cdot (2\theta + 3) \]

Differentiate \(\cos e^{\theta}\)

For \( \cos e^{\theta} \), treating \( e^{\theta} \) as an argument, let \( g(u) = \cos u \) and \( u = e^{\theta} \). Then:\[ \frac{d}{d\theta} \cos e^{\theta} = -\sin e^{\theta} \cdot e^{\theta} \]

Differentiate \(\sin(\frac{1}{\theta})\)

For \( \sin(\frac{1}{\theta}) \), let \( g(u) = \sin u \) and \( u = \frac{1}{\theta} \). The derivative is:\[ \frac{d}{d\theta} \sin(\frac{1}{\theta}) = \cos(\frac{1}{\theta}) \cdot \left(-\frac{1}{\theta^2}\right) \]

Summary of Derivatives

We have derived the derivatives for each function:- \( \frac{d}{d\theta} \cos \theta^3 = -3\theta^2 \sin \theta^3 \)- \( \frac{d}{d\theta} \sin(\theta^2 + 3\theta) = \cos(\theta^2 + 3\theta) (2\theta + 3) \)- \( \frac{d}{d\theta} \cos e^{\theta} = -e^{\theta} \sin e^{\theta} \)- \( \frac{d}{d\theta} \sin(1/\theta) = -\frac{\cos(1/\theta)}{\theta^2} \)

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Key Concepts

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

Derivative Formulas

Derivatives are tools used to compute rates of change. In calculus, the chain rule is essential when derivatives involve composite functions, which are functions within functions. To derive the derivative formulas for functions like \( \sin f(\theta) \) and \( \cos f(\theta) \), it's important to first understand the individual derivative formulas for sine and cosine:

The derivative of \( \sin x \) is \( \cos x \)
The derivative of \( \cos x \) is \( -\sin x \)

The chain rule helps us apply these basic derivatives when dealing with more complex functions like \( \sin f(\theta) \) and \( \cos f(\theta) \), making it possible to find their derivatives by identifying an inner and outer function. When we identify \( g(u) \), where \( u = f(\theta) \), the chain rule provides the necessary structure to compute these more intricate derivatives.

Composite Functions

Composite functions occur when the output of one function becomes the input of another. The chain rule is specifically useful in differentiating such functions. For instance, if \( y = \sin(f(\theta)) \), then \( \sin \) is the outer function and \( f(\theta) \) is the inner function. The chain rule links the derivative of the outer function to the derivative of the inner function:

Calculate the derivative of the outer function, \( \sin \, u \), as \( \cos u \)
Multiply by the derivative of the inner function, \( f(\theta) \)

So the derivative is \( \cos f(\theta) \cdot f'(\theta) \). By this method, composite polynomials, exponentials, and trigonometric functions can be dissected into manageable parts for differentiation.

Trigonometric Derivatives

Trigonometric derivatives are derivatives of trigonometric functions often combined with other types of functions, as seen through composite functions. Using trig identities is crucial here. Some of the basic trigonometric derivatives are:

\( \frac{d}{dx} \sin(x) = \cos(x) \)
\( \frac{d}{dx} \cos(x) = -\sin(x) \)
\( \frac{d}{dx} \tan(x) = \sec^2(x) \)

Utilizing the chain rule allows the calculation of derivatives like \( \frac{d}{d\theta} \sin(f(\theta)) \) by substituting the trigonometric derivative into the rule, creating a structured path to derive intricate expressions.

Calculus Techniques

Differentiating complex functions often involves combining multiple calculus techniques. The chain rule is a cornerstone, especially with composite functions. Using it involves:

Identifying the outer function and its derivative
Identifying the inner function and its derivative
Applying the chain rule: the derivative of the outer function evaluated at the inner function, then multiplied by the derivative of the inner function

Moreover, you might need to use other techniques such as the product rule for products of functions, or apply derivatives of exponential and logarithmic functions as well. Each technique interconnects, providing a complete toolbox to tackle a variety of differentiation problems.

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