/*! 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 64 Use logarithmic differentiation ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use logarithmic differentiation to evaluate $f^{\prime}(x)$$. $$f(x)=\frac{\tan ^{10} x}{(5 x+3)^{6}}$$

Short Answer

Expert verified
Question: Find the derivative of the given function \(f(x) = \frac{\tan ^{10} x}{(5 x+3)^{6}}\) using logarithmic differentiation. Answer: \(f'(x) = 10\tan^9 x\sec^2 x - 30\frac{\tan^{10}x}{(5x+3)^7}\)

Step by step solution

01

Take the natural logarithm of both sides

\(\ln(f(x)) = \ln\left(\frac{\tan ^{10} x}{(5 x+3)^{6}}\right)\).
02

Simplify the expression using logarithm properties

Use the properties of logarithms to simplify the right-hand side of the equation: \(\ln(f(x)) = 10\ln(\tan x) - 6\ln(5x+3)\).
03

Differentiate both sides of the equation with respect to x

Now, we'll differentiate both sides of the equation with respect to x: \(\frac{f'(x)}{f(x)} = 10\frac{d}{dx}(\ln(\tan x)) - 6\frac{d}{dx}(\ln(5x+3))\) Applying the chain rule, we get: \(\frac{f'(x)}{f(x)} = 10\frac{1}{\tan x}\cdot\frac{d}{dx}(\tan x) - 6\frac{1}{5x+3}\cdot\frac{d}{dx}(5x+3)\) \(\frac{f'(x)}{f(x)} = 10\frac{1}{\tan x}\cdot\sec^2 x - 6\frac{1}{5x+3}\cdot 5\)
04

Solve for \(f'(x)\), the derivative of the function

Now we can solve for \(f'(x)\): \(f'(x) = f(x)\cdot \left(10\frac{1}{\tan x}\cdot\sec^2 x - 6\frac{1}{5x+3}\cdot 5\right)\) We know that \(f(x) = \frac{\tan ^{10} x}{(5 x+3)^{6}}\). By multiplying both sides by the simplified expression, we get: \(f'(x) = \frac{\tan ^{10} x}{(5 x+3)^{6}}\cdot \left(10\frac{1}{\tan x}\cdot\sec^2 x - 6\frac{1}{5x+3}\cdot 5\right)\) Now simplify the expression: \(f'(x) = 10\tan^9 x\sec^2 x - 30\frac{\tan^{10}x}{(5x+3)^7}\) So the derivative of the given function is: \(f'(x) = 10\tan^9 x\sec^2 x - 30\frac{\tan^{10}x}{(5x+3)^7}\)

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 tool in calculus for differentiating composite functions. When you have a function nested inside another function, the chain rule helps you find the derivative. It states that if you have a composite function \( y = f(g(x)) \), then its derivative is the product of the derivative of the outer function and the derivative of the inner function. This can be mathematically represented as \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In the context of our exercise, when differentiating expressions such as \( \ln(\tan x) \), we apply the chain rule. The outer function here is the natural logarithm, and the inner function is \( \tan x \). Thus, the derivative \( \frac{d}{dx}(\ln(\tan x)) \) becomes \( \frac{1}{\tan x} \cdot \sec^2 x \). Similarly, for \( \ln(5x+3) \), the derivative is \( \frac{1}{5x+3} \cdot 5 \).

Applying the chain rule properly ensures accurate results especially in complex expressions involving multiple functions.
Derivative
A derivative represents the rate at which a function changes with respect to a variable. It's a fundamental concept in calculus, symbolized as \( f'(x) \) when differentiating a function \( f(x) \).
  • The derivative of a constant is zero.
  • The derivative of \( x^n \) is \( nx^{n-1} \).
  • The derivative of \( \sin x \) is \( \cos x \), and of \( \tan x \) is \( \sec^2 x \).
In logarithmic differentiation, we use the derivative of logarithmic functions, notably: \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).

Our problem involves differentiating a quotient \( \frac{\tan^{10} x}{(5x+3)^6} \). By taking the natural logarithm, simplifying using properties, and differentiating, we manage more complex differentiation using related derivatives like \( \tan x \) and the constants involved in polynomial terms. This approach breaks down the derivative into manageable parts, which are then combined for the final solution.
Logarithmic Properties
Logarithmic properties are crucial in simplifying complex expressions. Using logarithms in differentiation makes handling products, quotients, and powers more manageable.
  • The logarithm of a product is the sum of the logarithms: \( \ln(ab) = \ln a + \ln b \).
  • The logarithm of a quotient is the difference of the logarithms: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \).
  • The logarithm of a power brings down the exponent: \( \ln(a^b) = b\ln a \).
For our exercise, we first take the natural logarithm of both sides to leverage these properties. The function \( \frac{\tan^{10}x}{(5x+3)^6} \) becomes \( 10\ln(\tan x) - 6\ln(5x+3) \) after applying the properties. These simplified terms are easier to differentiate logarithmically, transforming a complex quotient into a manageable subtraction of two terms.

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. $$\cos \left(2 \sin ^{-1} x\right)=1-2 x^{2}$$

Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions \(\theta(t)\) and \(\varphi(t),\) respectively, where \(0 \leq t \leq 4\) and \(t\) is measured in minutes (see figure). These angles are measured in radians, where \(\theta=\varphi=0\) represent the starting position and \(\theta=\varphi=2 \pi\) represent the finish position. The angular velocities of the runners are \(\theta^{\prime}(t)\) and \(\varphi^{\prime}(t)\). a. Compare in words the angular velocity of the two runners and the progress of the race. b. Which runner has the greater average angular velocity? c. Who wins the race? d. Jean's position is given by \(\theta(t)=\pi t^{2} / 8 .\) What is her angular velocity at \(t=2\) and at what time is her angular velocity the greatest? e. Juan's position is given by \(\varphi(t)=\pi t(8-t) / 8 .\) What is his angular velocity at \(t=2\) and at what time is his angular velocity the greatest?

\(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that $$ \frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)} $$ b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0 ) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$$

a. Use derivatives to show that \(\tan ^{-1} \frac{2}{n^{2}}\) and \(\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\) differ by a constant. b. Prove that \(\tan ^{-1} \frac{2}{n^{2}}=\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\)

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.