Chapter 27: Problem 18
Find the derivatives of the given functions. $$y=\frac{7 \ln 3 x}{e^{2 x}+8}$$
Short Answer
Expert verified
The derivative is \( \frac{\left( \frac{7e^{2x}}{x} + \frac{56}{x} - 14e^{2x} \ln 3x \right)}{(e^{2x} + 8)^2} \).
Step by step solution
01
Identify the Derivative Rule
The function given is a quotient, so we'll use the quotient rule of derivatives. If you have a function \( y = \frac{u}{v} \), then the derivative \( y' \) is given by \( \frac{u'v - uv'}{v^2} \). Here, \( u = 7 \ln 3x \) and \( v = e^{2x} + 8 \).
02
Differentiate the Numerator
Let's differentiate the numerator \( u = 7 \ln 3x \). Using the derivative of a logarithm, \( \frac{d}{dx}[\ln u] = \frac{u'}{u} \), we have \( u'(3x) = \frac{3}{3x} \). Thus, \( u'(x) = 7 \times \frac{3}{3x} = \frac{7}{x} \).
03
Differentiate the Denominator
Now differentiate the denominator \( v = e^{2x} + 8 \). The derivative of \( e^{2x} \) is \( 2e^{2x} \) and the derivative of 8, a constant, is 0. So, \( v'(x) = 2e^{2x} \).
04
Apply the Quotient Rule
Substituting into the quotient rule: \( u = 7 \ln 3x \), \( u' = \frac{7}{x} \), \( v = e^{2x} + 8 \), \( v' = 2e^{2x} \), we get:\[\frac{d}{dx}[\frac{7 \ln 3x}{e^{2x} + 8}] = \frac{\left(\frac{7}{x}\right)(e^{2x} + 8) - 7 \ln 3x \times 2e^{2x}}{(e^{2x} + 8)^2}\]
05
Simplify the Expression
Simplify the expression:- Numerator: \( \frac{7}{x}(e^{2x} + 8) - 14e^{2x} \ln 3x \) - \( = 7 \frac{e^{2x}}{x} + \frac{56}{x} - 14e^{2x} \ln 3x \)- Denominator: \( (e^{2x} + 8)^2 \) So the derivative \( \frac{d}{dx}[\frac{7 \ln 3x}{e^{2x} + 8}] \) is:- \[ \frac{\left( \frac{7e^{2x}}{x} + \frac{56}{x} - 14e^{2x} \ln 3x \right)}{(e^{2x} + 8)^2} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
When you are dealing with a division of functions in calculus, such as \( \frac{u}{v} \), the quotient rule comes in handy to find the derivative. The core formula for the quotient rule is:\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\]Here, \( u \) and \( v \) correspond to the numerator and denominator functions, respectively. The expression involves taking the derivative of both \( u \) and \( v \), which are represented as \( u' \) and \( v' \).
- Differentiate the numerator: Apply rules like the product, chain, or power rule as needed. Identify the form of \( u \) and find its derivative.
- Differentiate the denominator: Similarly, identify the form of \( v \) and differentiate it.
- Substitute and simplify: Plug in the respective derivatives into the quotient rule formula and simplify to obtain your final result.
Logarithmic Differentiation
Logarithmic differentiation is a technique used when differentiating functions that involve logarithms, making it particularly useful when the function has products, quotients, or powers.In logarithmic differentiation, follow these steps:
- Take the natural logarithm of both sides of the equation, such as \( y = f(x) \). This transforms the original product, quotient, or power into a sum or difference, centralizing the logarithmic rules.
- Differentiate with respect to \( x \). Use the chain rule here, remembering that \( \frac{d}{dx}[\ln u] = \frac{u'}{u} \).
- Solve for the original derivative: After differentiating, multiply both sides by the original expression (if initially transformed) to solve for \( y' \).
Exponential Functions
Exponential functions, characterized by the equation \( f(x) = a^{bx+c} \), where \( a \) is a constant and \( b \) and \( c \) are parameters, appear frequently in both pure and applied mathematics. Differentiating exponential functions, particularly those with the natural base \( e \), follows a unique process:
- Basic Derivative: The derivative of \( e^{kx} \) with respect to \( x \) is \( ke^{kx} \), where \( k \) is a constant. This arises from the chain rule.
- Natural Base Properties: It's important to recall that the derivative of \( e^x \) is simply \( e^x \). This intrinsic feature of the exponential function often simplifies calculations when handling derivatives directly.
- Compound Exponential Functions: For products or quotients involving \( e^{ ext{yp}(x)} \), decompose the expression as needed to simplify differentiation with applicable rules like the product, quotient, or power rule.