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Magazine Chapter 3: Problem 4
Find \(d y / d x\). $$ y=\ln (2+\sqrt{x}) $$
Short Answer
Expert verified
The derivative is \(\frac{1}{(2 + \sqrt{x}) (2\sqrt{x})}\).
Step by step solution
01
Identify the outer function
The given function is of the form \(y = \ln(u)\), where \(u = 2 + \sqrt{x}\). The outer function is the natural logarithm, \(\ln(u)\).
02
Identify the inner function
The inner function is \(u = 2 + \sqrt{x}\), where the expression inside the natural logarithm includes a square root.
03
Differentiate the outer function
Differentiate \(f(u) = \ln(u)\) with respect to \(u\). The derivative is \(f'(u) = \frac{1}{u}\).
04
Differentiate the inner function
Differentiate \(u = 2 + \sqrt{x}\) with respect to \(x\). Recall that \(\sqrt{x} = x^{1/2}\). The derivative is \(u'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\).
05
Apply the chain rule
Apply the chain rule, which states \(\frac{dy}{dx} = f'(u) \cdot u'(x)\). Substitute \(f'(u) = \frac{1}{u}\) and \(u'(x) = \frac{1}{2\sqrt{x}}\) into the chain rule. Thus, \(\frac{dy}{dx} = \frac{1}{2 + \sqrt{x}} \cdot \frac{1}{2\sqrt{x}}\).
06
Simplify the expression
The expression becomes \(\frac{dy}{dx} = \frac{1}{(2 + \sqrt{x}) \cdot 2\sqrt{x}}\). This is the derivative of the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. When dealing with a function composed of two or more functions, the chain rule allows us to differentiate step by step. For example, in our problem, we have the composite function \( y = \ln(2+\sqrt{x}) \).
- Think about the function as consisting of two parts: the inner function \( u = 2 + \sqrt{x} \) and the outer function \( y = \ln(u) \).
- The key to applying the chain rule is to first differentiate the outer function with respect to the inner function: \( f'(u) = \frac{1}{u} \). Then, differentiate the inner function with respect to \( x \): \( u'(x) = \frac{1}{2\sqrt{x}} \).
- Finally, multiply these derivatives together: \( \frac{dy}{dx} = f'(u) \cdot u'(x) = \frac{1}{(2 + \sqrt{x})} \cdot \frac{1}{2\sqrt{x}} \).
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithm with base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. It鈥檚 particularly important in calculus because of its simplicity in differentiation and integration.
- The differentiation of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). This rule is pivotal when dealing with logarithmic differentiation.
- In our problem, \( y = \ln(2 + \sqrt{x}) \), the natural logarithm serves as the "outer function." By knowing its derivative is \( \frac{1}{u} \) when \( y = \ln(u) \), we can simplify the differentiation process using the chain rule.
- This property makes calculating derivatives of logarithmic functions straightforward, especially when combined with the chain rule for composite functions.
Differentiation
Differentiation involves finding the derivative of a function, which represents the rate of change of the function with respect to a variable. It is a core concept of calculus and is used for analyzing functions.
- In the function \( y = \ln(2 + \sqrt{x}) \), differentiation means finding \( \frac{dy}{dx} \), which tells us how \( y \) changes as \( x \) changes.
- We identify the inner and outer functions. Here, the inner function is \( u = 2 + \sqrt{x} \) and the outer function is \( y = \ln(u) \). We then differentiate these functions individually.
- For \( u = 2 + \sqrt{x} \), we write \( \sqrt{x} \) as \( x^{1/2} \) and differentiate: \( \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).
- Applying the chain rule, the derivative becomes \( \frac{dy}{dx} = \frac{1}{2 + \sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \), leading to the simplified result: \( \frac{1}{(2 + \sqrt{x}) \cdot 2\sqrt{x}} \).
