/*! 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 18 Derivative of In \(u\) Differe... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 18

Derivative of In \(u\) Differentiate. $$y=5.06 \ln \sqrt{x^{2}-3.25 x}$$

Short Answer

Expert verified
The derivative of the given function is \(y' = 2.53 \cdot \frac{1}{x^2 - 3.25x} \cdot (2x - 3.25)\).

Step by step solution

01

Rewrite the Function

Rewrite the natural logarithm of the square root as a power to simplify differentiation. Use the property \( \ln(x^n) = n \ln(x) \) to get \(y = 5.06 \ln((x^2 - 3.25x)^{1/2}) = 5.06 \cdot \frac{1}{2} \ln(x^2 - 3.25x)\).
02

Apply the Constant Multiple Rule

Differentiate the function using the constant multiple rule, which states that the derivative of a constant times a function is the constant times the derivative of the function. In this case, apply the constant \(2.53 = 5.06 \cdot \frac{1}{2}\) to the result of the derivative of the logarithmic part.
03

Differentiate the Logarithm

Differentiate the logarithmic function using the chain rule. The derivative of \( \ln(g(x)) \) is \(\frac{1}{g(x)}g'(x)\). Here \(g(x) = x^2 - 3.25x\), and its derivative \(g'(x)\) is \(2x - 3.25\).
04

Chain Rule Application

Combine the results of the previous steps to find the derivative of the original function. The final derivative of the function is \(y' = 2.53 \cdot \frac{1}{x^2 - 3.25x} \cdot (2x - 3.25)\).

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.

Chain Rule
Understanding the chain rule is critical in calculus, especially when differentiating composite functions. In essence, it provides a method for finding the derivative of a function that is composed of other functions. The chain rule states that the derivative of a composite function is the derivative of the outside function evaluated at the inside function, multiplied by the derivative of the inside function.

Natural Logarithm Differentiation
When dealing with the natural logarithm function, represented by \(\ln(x)\), its differentiation is quite straightforward. The basic derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\), given that \(x\) is a positive real number. The differentiation becomes slightly more involved when the argument of the logarithm is itself a function of \(x\), as in the provided exercise. This is where the chain rule plays an essential role in finding the derivative effectively.

Constant Multiple Rule
The constant multiple rule in differentiation is a handy tool when a function is multiplied by a constant. The rule simply states that the derivative of a constant times a function is the constant multiplied by the derivative of the function. Symbolically, for any constant \(a\) and function \(f(x)\), the derivative \(\frac{d}{dx}(a \(f\))\) equals \(a \(\frac{d}{dx}f\)\). This property makes it more convenient to handle expressions with coefficients, as seen in our exercise with the coefficient \(5.06\), before applying the differentiation of the logarithmic function.

Mathematical Properties of Logarithms
Logarithms have several mathematical properties that make them easier to work with, especially when simplifying expressions before differentiation. One of these properties allows you to convert a logarithm of an exponentiated argument into a product, that is, \(\ln(x^n) = n \ln(x)\). This was used in our exercise to simplify the square root within the logarithm before proceeding with the differentiation. Applying the properties of logarithms to rewrite expressions can often simplify the differentiation process and lead to more manageable results.

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

Find the horizontal distance \(\bar{x}\) to the centroid of the area formed by the curve \(y=\frac{1}{2}\left(e^{x}+e^{-x}\right),\) the coordinate axes, and the line \(x=1\)

Extreme Values and Points of Inflection Find the maximum, minimum, and inflection points for each curve. $$y=\ln \left(8 x-x^{2}\right)$$

Extreme Values and Points of Inflection Find the maximum, minimum, and inflection points for each curve. $$y=x \ln x$$

The height \(y\) of a certain pendulum released from a height of \(50.0 \mathrm{cm}\) is \(y=50.0 e^{-0.5 t} \mathrm{cm},\) where \(t\) is the time after release in seconds. Find the vertical component of the velocity of the pendulum when \(t=1.00 \mathrm{s}\).

Tangent to a Curve Find the slope of the tangent at the point indicated. $$y=\ln \left(x^{2}+2\right) \text { at } x=4$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

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.