/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 \(7 - 46\) Find the derivative o... [FREE SOLUTION] | 91Ó°ÊÓ

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\(7 - 46\) Find the derivative of the function. $$y = 2 ^ { \sin \pi x }$$

Short Answer

Expert verified
\(\frac{dy}{dx} = 2^{\sin \pi x} \cdot \pi \ln 2 \cdot \cos \pi x\)

Step by step solution

01

Identify the Function Type

The given function is an exponential function of the form \(y = a^{u(x)}\), where \(a = 2\) and \(u(x) = \sin \pi x\). To differentiate this type of function, we will need to use the chain rule together with the natural logarithm.
02

Apply Logarithmic Differentiation

Take the natural logarithm of both sides to make the process of differentiation easier:\[\ln y = \ln (2^{\sin \pi x})\]Using the logarithmic identity \(\ln (a^b) = b \ln a\), this becomes:\[\ln y = \sin \pi x \cdot \ln 2\]
03

Differentiate Both Sides

Differentiate both sides with respect to \(x\). Start by differentiating the left side:\[\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}\]Now differentiate the right side using the chain rule:\[\frac{d}{dx} (\sin \pi x \cdot \ln 2) = \ln 2 \cdot \frac{d}{dx}(\sin \pi x)\]
04

Differentiate \(\sin \pi x\)

Apply the chain rule to differentiate \(\sin \pi x\):\[\frac{d}{dx} (\sin \pi x) = \pi \cos \pi x\]
05

Combine Results

Substitute the results from Step 4 into the equation from Step 3:\[\frac{1}{y} \frac{dy}{dx} = \ln 2 \cdot \pi \cdot \cos \pi x\]
06

Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = y \cdot \pi \ln 2 \cdot \cos \pi x\]Substitute back \(y = 2^{\sin \pi x}\):\[\frac{dy}{dx} = 2^{\sin \pi x} \cdot \pi \ln 2 \cdot \cos \pi x\]
07

Simplify and Present Final Answer

The derivative \(\frac{dy}{dx}\) has been determined to be:\[\frac{dy}{dx} = 2^{\sin \pi x} \cdot \pi \ln 2 \cdot \cos \pi x\]

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Key Concepts

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

Exponential Functions
Exponential functions are a key concept in calculus and appear frequently in various mathematical contexts. These functions can be recognized by their basic form, where a constant base is raised to the power of a variable. The general expression of an exponential function is given as \( y = a^{u(x)} \), where \( a \) is the base and it is positive. In this specific problem, the base is \( 2 \), and the exponent is \( \sin \pi x \). Exponential functions are continuous and growing functions when the base is greater than 1, leading to applications in growth models and complex calculations.
The rapid rate of change of exponential functions makes their derivatives crucial in understanding the behavior of these functions, especially when involved in more complex expressions.
Logarithmic Differentiation
Logarithmic differentiation is a useful technique when dealing with functions that have variable exponents, like exponential functions. This method leverages the properties of logarithms to simplify the differentiation process. By taking the natural logarithm of both sides of an equation, we convert multiplication and powers into more manageable additions and multiplications. For example, the identity \( \ln(a^b) = b \ln(a) \) is particularly helpful, as it transforms the exponent into a coefficient.
In differentiating \( y = 2^{\sin \pi x} \), we used logarithmic differentiation to first simplify \( \ln y = \sin \pi x \ln 2 \), making it more straightforward to apply differentiation rules. This transformation allows us to deal with the exponent \( \sin \pi x \) separately, facilitating the differentiation task.
Chain Rule
The chain rule is an essential part of differentiation, particularly when a function is "nested" inside another function. It allows us to break down the differentiation process by considering the outer and the inner functions separately. The rule states that for a composite function \( y = f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).
In our problem, this concept applies when differentiating \( \ln(2) \cdot \sin \pi x \). We treat \( \sin \pi x \) as the inner function and differentiate it using the chain rule. Thus, \( \frac{d}{dx}(\sin \pi x) \) involves calculating the derivative of \( \sin x \) and applying a scale factor of \( \pi \) because of the \( \pi x \) term, resulting in \( \pi \cos \pi x \).
This step exemplifies the chain rule's power to handle derivatives of composed or complex functions, ensuring accuracy and ease.
Natural Logarithms
Natural logarithms, denoted by \( \ln \), are logarithms to the base \( e \), where \( e \) is approximately equal to 2.71828. They appear frequently in calculus due to their convenient properties when dealing with exponential functions. Natural logarithms are particularly useful because they simplify the derivative of exponential functions. When differentiating expressions involving \( e^x \), for instance, the derivative is simply \( e^x \), showing their neat relationship.
In the context of our differentiation problem, we apply natural logarithms to both sides of the equation \( y = 2^{\sin \pi x} \) to unfold the exponential property. This allows us to differentiate the resulting expression \( \ln y = \sin \pi x \cdot \ln 2 \) more easily by breaking down the complex exponent term and taking advantage of the linear relationship the logarithm introduces. Ultimately, this technique provides clarity and simplifies calculations that would otherwise be cumbersome.

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Most popular questions from this chapter

The frequency of vibrations of a vibrating violin string is given by $$\mathrm{f}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\rho}}$$ where \(L\) is the length of the string, T is its tension, and \(\rho\) is its linear density. [See Chapter 11 in D. E. Hall, Musical Acoustics, 3 \(\mathrm{d}\) ed. (Pacific Grove, CA: Brooks/Cole, \(2002 ) . ]\) (a) Find the rate of change of the frequency with respect to (i) the length (when T and \(\rho\) are constant), (ii) the tension (when L and \(\rho\) are constant), and (iii) the linear density (when L and T are constant). (b) The pitch of a note (how high or low the note sounds) is determined by the frequency f. (The higher the frequency, the higher the pitch.) Use the signs of the derivatives in part (a) to determine what happens to the pitch of a note (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates, (ii) when the tension is increased by turning a tuning peg. (iii) when the linear density is increased by switching to another string.

Find a formula for \(f^{(n)}(x)\) if \(f(x)=\ln (x-1)\)

On page 431 of Physics: Calculus, 2\(d\) ed, by Eugene Hecht (Pacific Grove, CA: Brooks/Cole, \(2000 ),\) in the course of deriving the formula \(\mathrm{T}=2 \pi \sqrt{\mathrm{L} / \mathrm{g}}\) for the period of a $$ \begin{array}{l}{\text { pendulum of length } L, \text { the author obtains the equation }} \\ {\mathrm{a}_{\mathrm{T}}=-g \sin \theta \text { for the tangential acceleration of the bob of the }}\end{array} $$ pendulum, He then says, "for small angles, the value of \(\theta\) in radians is very nearly the value of \(\sin \theta\) ; they differ by less than 2\(\%\) out to about \(20^{\circ} .\) $$ \begin{array}{c}{\text { (a) Verify the linear approximation at } 0 \text { for the sine function: }} \\ {\sin x=x}\end{array} $$ (b) Use a graphing device to determine the values of \(x\) for which sin \(x\) and \(x\) differ by less than 2\(\%\) . Then verify Hecht's statement by converting from radians to degrees.

Find the derivative. Simplify where possible. $$f(t)=\operatorname{sech}^{2}\left(e^{t}\right)$$

A television camera is positioned 4000 \(\mathrm{ft}\) from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 600 \(\mathrm{ft} / \mathrm{s}\) when it has risen 3000 \(\mathrm{ft}\) . (a) How fast is the distance from the television camera to the rocket changing at that moment? (b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?

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