/*! 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 22 Differentiate. \(f(x)=e^{\sin ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate. \(f(x)=e^{\sin 2 x}\).

Short Answer

Expert verified
The derivative of the function \(f(x) = e^{\sin 2x}\) is \(f'(x) = 2e^{\sin 2x} \cos 2x\).

Step by step solution

01

Identify the two functions within the composition

We need to identify the two functions that make up the composition: 1. The outer function, which is the exponential function: \(g(x) = e^x\) 2. The inner function, which is the sine function: \(h(x) = \sin 2x\) It's important to differentiate both these functions before applying the chain rule.
02

Differentiate the outer function

Differentiate the exponential function \(g(x) = e^x\) with respect to x: \[\frac{dg(x)}{dx} = e^x\]
03

Differentiate the inner function

Differentiate the sine function \(h(x) = \sin 2x\) with respect to x. We will need to use the chain rule again since it involves another composition of two functions: the sine function \(\sin x\) and the linear function \(2x\): Differentiate \(h(x) = \sin (2x)\) with respect to x: \[\frac{dh(x)}{dx} = \cos (2x) \cdot 2\]
04

Apply the chain rule

Now that we have differentiated both the inner and outer functions, we can apply the chain rule to find the derivative of the overall function \(f(x) = e^{\sin 2x}\): Using the chain rule, \[f'(x) = g'(h(x)) \cdot h'(x)\] Substituting values from Step 2 and Step 3, we get: \[f'(x) = e^{\sin 2x} \cdot (2 \cos 2x)\]
05

Simplify the result

The derivative of the function is already simplified. So the final answer is: \[f'(x) = 2e^{\sin 2x} \cos 2x\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus, especially crucial for differentiating composite functions. These are functions where one is nested inside another. Imagine peeling an orange: the peel is the outer function, and the juicy segments are the inner function. The chain rule helps us differentiate this peel and segments in terms of one variable. To use the chain rule, you need:
  • An outer function, like the peel or exponential function in math terms.
  • An inner function, like the segments or a sine function.
When you differentiate a composite function, you differentiate the outer function first, then multiply by the derivative of the inner function. If you think of it as helping one friend peel the orange while another friend segments it, that's teamwork! In math, you introduce the concept of function composition and call this process the chain rule. By breaking down the functions individually and applying them together, you get a well-rounded derivative of the composite function.
Exponential Function
The exponential function is a fascinating mathematical concept often denoted by the symbol \(e\). It's known for its amazing property where its rate of growth at any point is equal to its current value. In the context of differentiation, if you have \(f(x) = e^x\), then its derivative is also \(e^x\). This function models continuous growth or decay processes, like population growth or radioactive decay. You often work with exponentials when dealing with compounded continuous growth. This constancy in the derivative shows why the exponential function has such a unique role in calculus. Understanding this property means that differentiating more complex versions, like \(e^{ ext{sine function}}\), becomes achievable by using the chain rule.
Sine Function
The sine function is one of the most famous periodic functions in mathematics, frequently encountered in trigonometry. Imagine waves on an ocean: up and down in a rhythmic pattern, much like the sine wave you've seen graphed. It's a function dependent on angles, usually originating with respect to a circle and represented as \(\sin x\).
When differentiating the sine function, its derivative is \(\cos x\). When the sine function involves transformations like \(\sin 2x\), it stretches or compresses the wave, affecting its frequency. Applications of sine functions often appear in physics, notably in wave and oscillation studies. In calculus, knowing how to differentiate such functions prepares you for more complex problem solving, especially when they're part of a composite function differentiated with the help of the chain rule.
Calculus Problem Solving
Calculus is like the Swiss Army knife of mathematics, solving problems across various real-world scenarios. It deals with change and rates of change, like how fast a car speeds up. Problem solving in calculus, on the surface, seems challenging, but it's all about recognizing patterns and using tools like the chain rule.
In our original problem, differentiating \(f(x) = e^{\sin 2x}\) required identifying components and developing a plan. This plan involved breaking down the composite function with the chain rule, separately handling the sine function and the exponential function. It's like approaching a physics problem: understanding the pieces ensures accurate assembly of the solution. As you tackle more problems, the interplay between these functions like exponential and sine becomes clearer, helping simplify complex solutions in calculus. This guide of step-by-step ways will boost your calculus problem-solving confidence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\Omega\) be the region below the graph of \(y=e^{x}\) from \(x=0\) 10 \(x=1\) (a) Find the volume of the solid generated by revolving \(1 \mathrm{g} \Omega\) about the \(x\) -axis. (b) Set up the definite integral that gives the volume of the solid generated by revolving \(\Omega\) about the \(y\) -axis using the shell method. (You will see how to evaluate this integral in Section \(8.2 .)\)

Show that the average slope of the logarithm curve from \(x \quad a\) to \(x=b\) is $$\frac{1}{b-a} \ln \left(\frac{b}{a}\right)$$

The method of carbon dating makes use of the fact that all living organisms contain two isotopes of carbon, carbon- 12, denoted \(^{12} \mathrm{C}\) (a stable isotope), and carban-14, denoted \(^{14} \mathrm{C}\) (a radioactive isotope). The ratio of the amount of \(^{14} \mathrm{C}\) to the amount of \(^{12} \mathrm{C}\) is essentially constant (approximately \(1 / 10,000\) ). When an organism dies, the amount of \(^{12}\)C Present remains unchanged, but the \(^{14} \mathrm{C}\) decays at a rate proportional to the amount present with a half-life of approximately 5700 years. This change in the amount of \(^{14}\) Relative to the amount of \(^{12} \mathrm{C}\) makes it possible to estimate the time at which the organism lived. A fossil found in an archaeological dig was found to contain \(25 \%\) of the original amount of \(^{14} \mathrm{C}\). What is the approximate age of the fossil?

A particle moves along a coordinate line with acceleration \(a(t)=4-2(t+1)+3 /(t+1)\) feet per second per second from \(t=0\) to \(t=3\) (a) Find the velocity \(v\) of the particle at each time \(t\) during the motion given that \(v(0)=2\) (b) Use a graphing utility to graph \(v\) and \(a\) together. (c) Estimate the time \(t\) at which the particle has maximum velocity and the time at which it has minimum velocity. Use four decimal place accuracy.

Sketch the region bounded by the curves and find its area. \(y=e^{x}, \quad y=e, \quad y=x, \quad x=0\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.