/*! 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 38 Use the General Power Rule where... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 38

Use the General Power Rule where appropriate to find the derivative of the following functions. $$s(t)=\cos 2^{t}$$

Short Answer

Expert verified
Question: Find the derivative of the function $$s(t) = \cos 2^t$$ with respect to t. Answer: The derivative of the given function is $$\frac{ds}{dt} = -ln(2)2^t \sin(2^t)$$

Step by step solution

01

Identify the outer and inner functions

We first need to identify the outer function and the inner function of our composite function. In our case: Outer function: $$g(u) = \cos u$$ Inner function: $$u(t) = 2^t$$
02

Differentiate the outer function

We will now find the derivative of the outer function with respect to u: $$g'(u) = -\sin u$$
03

Differentiate the inner function

Next, we'll find the derivative of the inner function with respect to t. We will use the General Power Rule for this. $$u(t) = 2^t$$ Let $$v = 2$$ and $$du/dt = ln(2)2^{t-1}$$ (this comes from the General Power Rule). The derivative of the inner function is: $$\frac{du}{dt} = ln(2)2^t$$
04

Apply the Chain Rule

Now that we have the derivatives of both the outer and inner functions, we can apply the chain rule: $$\frac{ds}{dt} = g'(u(t)) \cdot \frac{du}{dt}$$ Substitute the derivatives we found in Steps 2 and 3: $$\frac{ds}{dt} = -(sin(2^t)) \cdot (ln(2)2^t)$$ So, the derivative of the given function is: $$\frac{ds}{dt} = -ln(2)2^t \sin(2^t)$$

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 91影视!

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

Derivatives from tangent lines Suppose the line tangent to the graph of \(f\) at \(x=2\) is \(y=4 x+1\) and suppose \(y=3 x-2\) is the line tangent to the graph of \(g\) at \(x=2 .\) Find an equation of the line tangent to the following curves at \(x=2\) a. \(y=f(x) g(x)\) b. \(y=\frac{f(x)}{g(x)}\)

Find the derivative of the inverse of the following functions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\) $$f(x)=x^{2}-2 x-3, \text { for } x \leq 1 ;(12,-3)$$

Calculate the following derivatives using the Product Rule. $$\begin{array}{lll} \text { a. } \frac{d}{d x}\left(\sin ^{2} x\right) & \text { b. } \frac{d}{d x}\left(\sin ^{3} x\right) & \text { c. } \frac{d}{d x}\left(\sin ^{4} x\right) \end{array}$$ d. Based upon your answers to parts (a)-(c), make a conjecture about \(\frac{d}{d x}\left(\sin ^{n} x\right),\) where \(n\) is a positive integer. Then prove the result by induction.

Use any method to evaluate the derivative of the following functions. $$f(x)=4 x^{2}-\frac{2 x}{5 x+1}$$

The hands of the clock in the tower of the Houses of Parliament in London are approximately \(3 \mathrm{m}\) and \(2.5 \mathrm{m}\) in length. How fast is the distance between the tips of the hands changing at 9:00? (Hint: Use the Law of cosines.)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation
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.