/*! 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 Differentiate the function. \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 31

Differentiate the function. \(G(y)=\frac{A}{y^{10}}+B e^{y}\)

Short Answer

Expert verified
The derivative is \( G'(y) = -10 A y^{-11} + B e^{y} \).

Step by step solution

01

Identify the Terms

The function is given as \(G(y) = \frac{A}{y^{10}} + B e^{y}\). The first term is \(\frac{A}{y^{10}}\) and the second term is \(B e^{y}\). We will differentiate each term separately.
02

Differentiate the First Term

Rewrite the first term \(\frac{A}{y^{10}}\) as \(A y^{-10}\). Apply the power rule for differentiation: \( \frac{d}{dy} y^n = n y^{n-1} \). Thus, \( \frac{d}{dy} (A y^{-10}) = -10 A y^{-11} \).
03

Differentiate the Second Term

The second term \(B e^{y}\) has the derivative of \(B e^{y}\) because the derivative of \(e^y\) with respect to \(y\) is \(e^y\). Thus, \( \frac{d}{dy} (B e^{y}) = B e^{y} \).
04

Combine the Derivatives

Combine the results from Steps 2 and 3 to find the derivative of the original function. This gives \( G'(y) = -10 A y^{-11} + B e^{y} \).

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!

Key Concepts

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

Power Rule Differentiation
The power rule is a fundamental concept in calculus for finding derivatives. It simplifies the process when differentiating functions of the form \(y^n\). In this exercise, we used the power rule to differentiate \(\frac{A}{y^{10}}\) by rewriting it as \(A y^{-10}\). This allows us to apply the power rule directly.
  • The power rule states: \(\frac{d}{dy} y^n = n y^{n-1}\).
  • The exponent \(n\) is brought down as a coefficient in front of the term.
  • The new exponent becomes \(n-1\).

For our specific term, \(A y^{-10}\), applying the rule gives us:
  • First, bring \(-10\) as a coefficient: \(-10 A\).
  • Then, subtract one from the exponent: \(y^{-10-1} = y^{-11}\).
This results in \(-10 A y^{-11}\), the derivative of \(A y^{-10}\).
This step shows how efficiently the power rule works, turning difficult-looking fractions into manageable expressions.
Exponential Function Differentiation
Exponential functions, where the variable is in the exponent, such as \(e^y\), have their own rule for differentiation. The beauty of these functions is that their derivatives are surprisingly simple to handle.For the exponential term \(B e^y\) from our exercise, we follow the basic differentiation rule for \(e^y\):
  • The derivative of \(e^y\) is just \(e^y\) again.
  • When a constant \(B\) is multiplied with \(e^y\), this constant remains in the differentiated term.

Thus, the differentiation of \(B e^y\) is straightforward:
  • \(\frac{d}{dy} (B e^y) = B \cdot e^y = B e^y\).
The simplicity of exponential function differentiation makes exponential terms predictable and easy to manage in calculus.
Step-by-Step Differentiation Process
When tackling a differentiation problem, breaking it down into manageable steps is key. This process ensures clarity and accuracy. Let鈥檚 go over the steps used in our exercise to differentiate \(G(y) = \frac{A}{y^{10}} + B e^y\).First, identify all terms of the function:
  • Recognize that the function consists of two distinct parts: \(\frac{A}{y^{10}}\) and \(B e^y\).

Next, perform differentiation on each term separately:
  • For \(\frac{A}{y^{10}}\): Rewrite it as \(A y^{-10}\) and apply the power rule to get \(-10 A y^{-11}\).
  • For \(B e^y\): Recognize it as an exponential function and apply the exponential differentiation rule to get \(B e^y\).

Finally, combine the differentiated terms:
  • Add the results: \(-10 A y^{-11} + B e^y\).
This structured, step-by-step approach to differentiation simplifies complex expressions and minimizes errors, making calculus problems more approachable.

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

(a) Use the definition of the derivative to calculate \(f^{\prime}\) , (b) Check to see that your answer is reasonable by comparing the graphs of \(f\) and \(f^{\prime}\) \(f(t)=t^{2}-\sqrt{t}\)

Find the derivative of the function using the definition of a derivative. State the domain of the function and the domain of its derivative. \(f(x)=1.5 x^{2}-x+3.7\)

\begin{array}{c}{\text { (a) If } g \text { is differentiable, the Reciprocal Rule says that }} \\ {\frac{d}{d x}\left[\frac{1}{g(x)}\right]=-\frac{g^{\prime}(x)}{[g(x)]^{2}}} \\ {\text { Use the Reciprocal Rule to prove the Reciprocal Rule. }} \\ {\text { (b) Use the Reciprocal Rule to differentiate the function. }}\\\ {y=1 /\left(x^{4}+x^{2}+1\right)} \\ {\text { (c) Use the Reciprocal Rule to verify that the Power Rule }} \\ {\text { is valid for negative integers, that is, }}\\\ \frac{d}{d x}\left(x^{-n}\right)=-n x^{-n-1}\\\ {\text { for all positive integers \(n\). }} \end{array}

Use the Chain Rule to show that if \(\theta\) is measured in degrees, then \(\frac{d}{d \theta}(\sin \theta)=\frac{\pi}{180} \cos \theta\) (This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: The differentiation formulas would not be as simple if we used degree measure.)

\(43-48\) Find the derivative of the function. Simplify where possible. $$y=\left(\tan ^{-1} x\right)^{2}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

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.