Chapter 3: Problem 12
Give the derivative formula for each function. \(\quad j(x)=-\ln x+\frac{1}{2 e^{4}}\)
Short Answer
Expert verified
The derivative is \( j'(x) = -\frac{1}{x} \).
Step by step solution
01
Identify the Function Components
The function given is \( j(x) = -\ln x + \frac{1}{2 e^4} \). It consists of two parts: the term \(-\ln x\) and the constant term \(\frac{1}{2 e^4}\).
02
Differentiate the First Term
Differentiate the term \(-\ln x\) with respect to \(x\). The derivative of \(\ln x\) is \(\frac{1}{x}\), so the derivative of \(-\ln x\) is \(-\frac{1}{x}\).
03
Differentiate the Constant Term
The derivative of a constant is 0, and since \(\frac{1}{2e^4}\) is a constant, its derivative is 0.
04
Combine the Derivatives
Combine the derivatives from Step 2 and Step 3: \(-\frac{1}{x} + 0\). Simplifying this gives us \(-\frac{1}{x}\).
05
State the Derivative Formula
The derivative \( j'(x) \) of the function \( j(x) = -\ln x + \frac{1}{2e^4} \) is \( j'(x) = -\frac{1}{x} \).
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.
Natural Logarithm
Natural logarithms play a fundamental role in calculus and derive from the function \( y = \ln x \). The natural log represents the exponent to which the base \( e \) (approximately 2.718) must be raised to obtain a given number. In simple terms, \( \ln x \) answers the question: "What power should we raise \( e \) to, in order to get \( x \)?" It's quite useful in many fields like finance and biology because of its continuous and smooth nature.
When dealing with derivatives, the natural logarithm follows a straightforward rule. For a function \( f(x) = \ln x \), the derivative is \( f'(x) = \frac{1}{x} \). This tells us how the logarithmic function changes with respect to \( x \). If multiplied by negative ones like in \( -\ln x \), the derivative becomes \( -\frac{1}{x} \). The minus sign indicates the slope of the curve is inverted, making it decrease rather than increase.
When dealing with derivatives, the natural logarithm follows a straightforward rule. For a function \( f(x) = \ln x \), the derivative is \( f'(x) = \frac{1}{x} \). This tells us how the logarithmic function changes with respect to \( x \). If multiplied by negative ones like in \( -\ln x \), the derivative becomes \( -\frac{1}{x} \). The minus sign indicates the slope of the curve is inverted, making it decrease rather than increase.
Calculus Differentiation
Differentiation in calculus is the process of finding the derivative of a function. The derivative provides a formula for the rate at which a function is changing at any given point. This is particularly useful in understanding curves and slopes.
- The function \( j(x) = -\ln x + \frac{1}{2e^4} \) needs to be differentiated term by term.
- When you take the derivative of a composite function like this, consider each part separately.
- The differentiation of \( -\ln x \) is straightforward as explained earlier: \(-\frac{1}{x}\).
Constant Function
In calculus, constant functions are quite simple to handle because they always differentiate to zero. A constant function is one where the output value remains the same regardless of the input. For example, \( f(x) = c \) where \( c \) is a constant such as \( \frac{1}{2e^4} \).
- When differentiating a constant, remember the rule that its derivative is zero because the rate of change of a constant is non-existent.
- This allows us to focus on the variable parts of a function during differentiation.