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Find the derivatives of the functions. $$s=\frac{4}{3 \pi} \sin 3 t+\frac{4}{5 \pi} \cos 5 t$$

Short Answer

Expert verified
The derivative is \( \frac{4}{\pi} \cos 3t - \frac{4}{\pi} \sin 5t \).

Step by step solution

01

Identify the Derivative Rules

To find the derivative of the function, we will use the derivative rules for sine and cosine. The derivative of \( \sin(kx) \) is \( k\cos(kx) \), and the derivative of \( \cos(kx) \) is \( -k\sin(kx) \).
02

Apply the Derivative to Each Term

The function is \( s = \frac{4}{3 \pi} \sin 3t + \frac{4}{5 \pi} \cos 5t \). We apply the derivative rules separately to each term. For \( \frac{4}{3 \pi} \sin 3t \), its derivative is \( \frac{4}{3 \pi} \times 3 \cos 3t = \frac{12}{3 \pi} \cos 3t \). For \( \frac{4}{5 \pi} \cos 5t \), its derivative is \( -\frac{4}{5 \pi} \times 5 \sin 5t = -\frac{20}{5 \pi} \sin 5t \).
03

Simplify the Expressions

Simplify the derivatives we found: \( \frac{12}{3 \pi} \cos 3t = \frac{4}{\pi} \cos 3t \) and \( -\frac{20}{5 \pi} \sin 5t = -\frac{4}{\pi} \sin 5t \).
04

Write the Final Derivative Function

Combine the simplified derivatives to form the derivative of the function \( s \). Therefore, \( \frac{ds}{dt} = \frac{4}{\pi} \cos 3t - \frac{4}{\pi} \sin 5t \).

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Key Concepts

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

Trigonometric Functions
Trigonometric functions like sine and cosine are foundational in calculus and trigonometry. They describe the properties of angles and the relationships in physical waves. When you differentiate these functions, you uncover rates of change or slopes related to oscillating phenomena. The sine function, denoted as \( \sin x \), depicts a smooth wave-like curve that repeats every \( 2\pi \). The cosine function, \( \cos x \), has a similar curve that starts at its peak. Both functions are periodic and their derivatives possess unique properties, which continually feed into their cyclical nature.

In the context of derivatives:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \(-\sin x \).
Given this, differentiating a sine or cosine function is like finding a new function that matches the original function’s rate of change at every point along its curve. Such insights are invaluable in understanding wave motion and harmonic oscillations.
Derivative Rules
Differentiation rules enable us to calculate the rates of change of complex functions. These rules simplify the calculus process by setting guidelines to manage derivatives efficiently. Particularly for trigonometric functions, specific rules link their cyclic nature to their derivatives.

Key derivative rules include:
  • Power rule links terms like \( x^n \) to derivatives of \( nx^{n-1} \).
  • Product rule aids in differentiating products of two functions.
  • Quotient rule helps tackle the divisions of functions.
  • Chain rule deals with composed or nested functions.
In our problem, recognizing the trigonometric derivative rules specifically for \( \sin(kx) \) and \( \cos(kx) \) was essential. Here:
  • \( \frac{d}{dx} \sin(kx) = k \cos(kx) \)
  • \( \frac{d}{dx} \cos(kx) = -k \sin(kx) \)
These rules allowed us to apply a simple multiplier from the coefficient of \( k \), showing just how powerful these rules are in making complex derivatives much more manageable.
Chain Rule
The chain rule in calculus is a method for differentiating composite functions. It's important when a function is nested inside another. For example, a function of the form \( f(g(x)) \). Understanding and applying the chain rule leads to finding the derivative by multiplying the derivative of the outer function by the derivative of the inner function.

In our task of differentiating \( \frac{4}{3 \pi} \sin 3t + \frac{4}{5 \pi} \cos 5t \):
  • \( \sin 3t \) and \( \cos 5t \) are the nested inner functions \( g(x) = 3t \) and \( g(x) = 5t \), each with their outer trigonometric functions.
  • The chain rule impacts derivatives as \( 3 \) and \( 5 \) are multiplied to the respective derivatives of the trigonometric functions.
Applying the chain rule returns derivatives formatted as \( \frac{4}{3 \pi} \times 3 \cos 3t \) and \( \frac{4}{5 \pi} \times (-5) \sin 5t \). This illustration of the chain rule portrays the intuitive extension required for nested functions. Learning to apply the chain rule equips you with a tool to tackle unusual situations where functions overlap.

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Most popular questions from this chapter

Find \(d y / d x\) $$\ln y=e^{y} \sin x$$

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=\sqrt{3 x-2}, \quad \frac{2}{3} \leq x \leq 4, \quad x_{0}=3$$

What is special about the functions $$f(x)=\sin ^{-1} \frac{1}{\sqrt{x^{2}+1}} \text { and } g(x)=\tan ^{-1} \frac{1}{x} ?$$ Explain.

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=e^{x}, \quad-3 \leq x \leq 5, \quad x_{0}=1$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{5} e^{x}$$

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