/*! 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 26 Differentiate. $$ y=6^{x} \c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate. $$ y=6^{x} \cdot \log _{7} x $$

Short Answer

Expert verified
The derivative is \( y' = 6^{x} \cdot \ln(6) \cdot \log_{7} x + \frac{6^{x}}{x \cdot \ln 7} \).

Step by step solution

01

Apply Product Rule

To differentiate the function \( y = 6^{x} \cdot \log_{7} x \), we need to use the product rule. The product rule states that if \( y = u \cdot v \), then \( y' = u' \cdot v + u \cdot v' \). Here, \( u = 6^{x} \) and \( v = \log_{7} x \).
02

Differentiate \( u = 6^{x} \)

To differentiate \( 6^{x} \), use the formula for differentiating exponential functions: \( \frac{d}{dx} a^{x} = a^{x} \cdot \ln(a) \). Hence, \( \frac{d}{dx} 6^{x} = 6^{x} \cdot \ln(6) \).
03

Differentiate \( v = \log_{7} x \)

To differentiate \( \log_{7} x \), first express it in terms of the natural logarithm: \( \log_{7} x = \frac{\ln x}{\ln 7} \). Using the formula \( \frac{d}{dx} \ln x = \frac{1}{x} \), we get \( \frac{d}{dx} \log_{7} x = \frac{1}{x \cdot \ln 7} \).
04

Substitute into the Product Rule

Replace \( u' \), \( u \), \( v' \), and \( v \) in the product rule formula. So, \( y' = (6^{x} \cdot \ln(6)) \cdot \log_{7} x + 6^{x} \cdot (\frac{1}{x \cdot \ln 7}) \).
05

Simplify the Expression

Combine and simplify the expression: \[ y' = (6^{x} \cdot \ln(6) \cdot \log_{7} x) + \left( \frac{6^{x}}{x \cdot \ln 7} \right) \]. This is the derivative of the original function.

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.

Product Rule
Let's begin with the product rule, an essential tool in differential calculus. When you have a function that is a product of two other functions, such as \( y = u \cdot v \), the derivative is not as straightforward as taking the derivative of each separately. Instead, we apply the product rule. The product rule states:
  • If \( y = u \cdot v \), then the derivative \( y' = u' \cdot v + u \cdot v' \).
This can be summarized as "derivative of the first times the second plus the first times the derivative of the second." If you remember only this structure, you will handle many differentiation problems with ease. By using the product rule, we ensure that both parts of a product function are properly accounted for when finding the derivative. It's particularly helpful when differentiating complex functions that can't be simplified into sums or differences.
Exponential Functions
Exponential functions are powerful and are of the form \( a^x \), where \( a \) is a constant and \( x \) is an exponent. These functions grow very rapidly and their differentiation involves a special rule. The derivative of an exponential function \( a^x \) is given by: \[\frac{d}{dx} a^x = a^x \cdot \ln(a)\] Here, \( \ln(a) \) is the natural logarithm of the base \( a \).
  • For instance, for \( 6^x \), the derivative is \( 6^x \cdot \ln(6) \).
  • This formula is derived from the fact that exponential functions are closely tied to their own rates of growth.
Exponential functions are not only found in pure math problems but also in real-world applications like population growth, radioactive decay, and financial calculations. Understanding these proportions makes applying their derivatives intuitive.
Logarithmic Differentiation
Logarithmic differentiation is often used to differentiate complicated functions that involve logarithms. One special case is when dealing with functions like \( \log_{b} x \), which can be tricky without leveraging natural logarithms (\( \ln \)). For the differentiation of \( \log_{b} x \), it's useful to rewrite it as \( \frac{\ln x}{\ln b} \). Then, using the differentiation rule for natural logs: \[\frac{d}{dx} \ln x = \frac{1}{x}\] Applying this gives: \[\frac{d}{dx} \log_{b} x = \frac{1}{x \cdot \ln b}\]
  • This approach effectively simplifies the process of differentiating with respect to \( x \).
  • It's particularly useful when the logarithmic base is other than \( e \), the base of natural logarithms.
Logarithmic differentiation offers a streamlined method to deal with otherwise complicated derivatives, especially when combined with other differentiation rules such as the product or chain rules.

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

The value, \(V(t),\) in dollars, of a share of Cypress Mills stock \(t\) months after it is purchased is modeled by $$ V(t)=58\left(1-e^{-1.1 t}\right)+20$$. a) Find \(V(1)\) and \(V(12)\). b) Find \(V^{\prime}(t)\). c) After how many months will the value of a share of the stock first reach \(\$ 75 ?\) d) Find \(\lim _{t \rightarrow \infty} V(t)\). Discuss the value of a share over a long period of time. Is this trend typical?

A beam of light enters a medium such as water or smoky air with initial intensity \(I_{0}\). Its intensity is decreased depending on the thickness (or concentration) of the medium. The intensity I at a depth (or concentration) of \(x\) units is given by $$I=I_{0} e^{-\mu x}$$ The constant \(\mu\left({ }^{u} m u "\right),\) called the coefficient of absorption, varies with the medium. Use this law for Exercises 62 and \(63 .\) Concentrations of particulates in the air due to pollution reduce sunlight. In a smoggy area, \(\mu=0.01\) and \(x\) is the concentration of particulates measured in micrograms per cubic meter \(\left(\mathrm{mcg} / \mathrm{m}^{3}\right)\) What change is more significant-dropping pollution levels from \(100 \mathrm{mcg} / \mathrm{m}^{3}\) to \(90 \mathrm{mcg} / \mathrm{m}^{3}\) or dropping them from \(60 \mathrm{mcg} / \mathrm{m}^{3}\) to \(50 \mathrm{mcg} / \mathrm{m}^{3}\) ? Why?

Differentiate. $$ f(x)=\log _{5} x $$

Differentiate. $$ f(x)=e^{\sqrt{x}}+\sqrt{e^{x}} $$

Pharmaceutical firms invest significantly in testing new medications. After a drug is approved by the Federal Drug Administration, it still takes time for physicians to fully accept and start prescribing it. The acceptance by physicians approaches a limiting value of \(100 \%,\) or \(1,\) after \(t\) months. Suppose that the percentage \(P\) of physicians prescribing a new cancer medication after \(t\) months is approximated by $$ P(t)=100\left(1-e^{-0.4 t}\right) $$ a) What percentage of doctors are prescribing the medication after 0 months? 1 month? 2 months? 3 months? 5 months? 12 months? 16 months? b) Find \(P^{\prime}(7)\), and interpret its meaning. c) Sketch a graph of the function.

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.