/*! 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 37 Use the General Power Rule where... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 37

Use the General Power Rule where appropriate to find the derivative of the following functions. $$g(y)=e^{y} \cdot y^{e}$$

Short Answer

Expert verified
Answer: The derivative of the function $$g(y) = e^{y} \cdot y^{e}$$ with respect to y is $$g'(y) = e^{y \cdot y^e} (y^e (e + e^y))$$.

Step by step solution

01

Identify u(y) and v(y)

Observe that a given function $$g(y) = e^{y} \cdot y^{e}$$ has the form $$u(y)^{v(y)}$$. Therefore, we can identify the functions $$u(y)$$ and $$v(y)$$ as: $$u(y) = e^y$$ $$v(y) = y^e$$
02

Calculate the derivatives of u(y) and v(y)

Now, we need to find the derivatives of both $$u(y)$$ and $$v(y)$$ with respect to y. Derivative of $$u(y) = e^y$$ with respect to y: $$u'(y) = \frac{d}{dy} (e^y) = e^y$$ Derivative of $$v(y) = y^e$$ with respect to y: $$v'(y) = \frac{d}{dy} (y^e) = e \cdot y^{e-1}$$
03

Apply the General Power Rule

Now that we have the derivatives of $$u(y)$$ and $$v(y)$$, we can apply the General Power Rule to find the derivative of the function $$g(y)$$: $$g'(y) = u(y)^{v(y)} (v'(y) \ln(u(y)) + v(y) u'(y))$$ Substitute the expressions found in steps 1 and 2 into this formula, and simplify: $$g'(y) = (e^y)^{y^e} (e \cdot y^{e-1} \ln(e^y) + y^e \cdot e^y)$$
04

Simplify the result

Simplify the expression for $$g'(y)$$ using the logarithmic property $$\ln(a^b) = b \ln(a)$$: $$g'(y) = e^{y \cdot y^e} (e \cdot y^{e-1} \cdot y \ln(e) + y^e \cdot e^y)$$ Since $$\ln(e) = 1$$, it further simplifies to: $$g'(y) = e^{y \cdot y^e} (e \cdot y^e + y^e \cdot e^y)$$ Now, the derivative of the function $$g(y)$$ is found: $$g'(y) = e^{y \cdot y^e} (y^e (e + e^y))$$

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.

General Power Rule
The General Power Rule in calculus provides a method for differentiating functions of the form \( u(y)^{v(y)} \). In this exercise, we looked at a function \( g(y) = e^y \, y^e \). To apply the General Power Rule, it is crucial to first identify functions \( u(y) \) and \( v(y) \) within the given expression.
  • Identify Components: Recognize \( g(y) \) as \( u(y)^{v(y)} \). For our exercise, \( u(y) = e^y \) and \( v(y) = y^e \).
  • Apply Rule: According to the rule, the derivative is calculated using:
    \[ g'(y) = u(y)^{v(y)} \bigg( v'(y) \ln(u(y)) + v(y) \frac{d}{dy}(u(y)) \bigg) \]
Understanding each variable and calculating its components separately can simplify complex differentiation problems, making it look less formidable. This ensures that we are methodically applying the formula to get the derivative effectively.
Derivative
Derivatives are fundamental in calculus, describing how a function changes as its input changes. In the context of this exercise, we want to find how the function \( g(y) \) changes with respect to \( y \).
  • Why Differentiate? Derivatives help us understand the rate of change. For \( g(y) = e^y \, y^e \), determine how each part of the function reacts to small changes in \( y \).
  • Calculate Derivatives: Compute derivatives of the individual components:
    - \( u(y) = e^y \), hence \( u'(y) = e^y \).
    - \( v(y) = y^e \), hence \( v'(y) = e y^{e-1} \).
The understanding of derivatives not only provides speed and direction but is essential for simplifying functions and tackling the task systematically.
Exponentiation
Exponentiation is a mathematical operation involving two numbers, the base and the exponent. In our function \( g(y) = e^y \, y^e \), both exponential expressions play a vital role.
  • Exponential Functions: Here, \( e^y \) is an exponential function with a constant base \( e \), offering steady growth with an increasing \( y \).
  • Variable Exponents: The expression \( y^e \) has a variable base \( y \) and a constant exponent \( e \), scaling differently based on the power given by \( e \).
Understanding the nature of exponential growth and behavior of base and exponents helps in differentiating adequately and recognizing patterns in the functions.
Logarithmic Properties
Logarithmic properties can simplify expressions, proving especially helpful in calculus. In this context, the logarithm helps in simplifying our derived expressions for\( g(y) \).
  • Properties Utilized: We used \( \ln(a^b) = b\ln(a) \), which is a property that transforms powers into multiplicative functions.
  • Application in Simplification: For instance, using \( \ln(e^y) = y\ln(e) \), we easily identify that the natural log of \( e \) equals \( 1 \), simplifying further calculations.
Grasping these properties is key, as they reduce seemingly complicated calculations into manageable steps and provide insight into the relationship between exponential and logarithmic processes.

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

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(\sin ^{-1} x\right)=\sqrt{1-x^{2}}$$

The volume of a torus (doughnut or bagel) with an inner radius of \(a\) and an outer radius of \(b\) is \(V=\pi^{2}(b+a)(b-a)^{2} / 4\) a. Find \(d b / d a\) for a torus with a volume of \(64 \pi^{2}\). b. Evaluate this derivative when \(a=6\) and \(b=10\)

A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance \(x\) (in inches) of the mass from its equilibrium position after \(t\) seconds is given by the function \(x(t)=10 \sin t-10 \cos t,\) where \(x\) is positive when the mass is above the equilibrium position. a. Graph and interpret this function. b. Find \(\frac{d x}{d t}\) and interpret the meaning of this derivative. c. At what times is the velocity of the mass zero? d. The function given here is a model for the motion of an object on a spring. In what ways is this model unrealistic?

a. Identify the inner function \(g\) and the outer function \(f\) for the composition \(f(g(x))=e^{k x},\) where \(k\) is a real number. b. Use the Chain Rule to show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\).

Find the following higher-order derivatives. $$\left.\frac{d^{3}}{d x^{3}}\left(x^{4.2}\right)\right|_{x=1}$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

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.