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Differentiate the function. \( g(t) = \sqrt {1 + \ln t} \)

Short Answer

Expert verified
The derivative is \( g'(t) = \frac{1}{2t \sqrt{1 + \ln t}} \).

Step by step solution

01

Rewrite the Function

First, rewrite the function in a form that is easier to differentiate. The given function is \( g(t) = \sqrt{1 + \ln t} \). This can be rewritten using a power of 1/2: \( g(t) = (1 + \ln t)^{1/2} \).
02

Apply the Chain Rule

To differentiate \( g(t) = (1 + \ln t)^{1/2} \), use the chain rule. The chain rule states that \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \). Here, set \( f(u) = u^{1/2} \) and \( u = 1 + \ln t \).
03

Differentiate the Outer Function

Differentiate the outer function \( f(u) = u^{1/2} \) with respect to \( u \). The derivative is \( f'(u) = \frac{1}{2} u^{-1/2} \).
04

Differentiate the Inner Function

Differentiate the inner function \( u = 1 + \ln t \) with respect to \( t \). The derivative is \( u'(t) = \frac{d}{dt}(1 + \ln t) = \frac{1}{t} \).
05

Combine the Derivatives Using the Chain Rule

Combine the derivatives from Steps 3 and 4 using the chain rule. This gives: \[ g'(t) = \frac{1}{2} (1 + \ln t)^{-1/2} \cdot \frac{1}{t} \].
06

Simplify the Expression

Simplify the resulting expression for the derivative: \[ g'(t) = \frac{1}{2t \sqrt{1 + \ln t}} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Chain Rule in Differentiation
The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. A composite function is essentially a function within another function. To grasp the idea of the chain rule, imagine peeling an onion layer by layer, until you get to the core. Each layer needs to be handled with care, symbolically represented by different expressions in differential calculus.
For a composite function like \( f(g(t)) \), the chain rule tells us how to find the derivative. It states: \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \). Here, you first differentiate the outer function \( f \) with respect to the inner function \( g \), and then multiply this result by the derivative of the inner function \( g \) itself.
  • This rule allows us to effectively "untangle" these composite structures and find their rates of change.
  • It is especially useful for functions that are raised to powers, contain exponential terms, or include logarithmic elements.
This example uses the chain rule to differentiate a function of the form \( g(t) = (1 + \ln t)^{1/2} \), resulting in a combination of derivatives applied consecutively. This structured approach simplifies otherwise complex differentiation tasks.
Differentiation of Natural Logarithm
When differentiating expressions with natural logarithms, it's crucial to understand the properties of the logarithmic function. The natural logarithm, denoted as \( \ln(x) \), is a common logarithm with base \( e \), where \( e \approx 2.718 \). This logarithm has a unique property that simplifies its differentiation significantly.
The derivative of the natural logarithm \( \ln(t) \) with respect to \( t \) is specially defined as \( \frac{1}{t} \). This assumes that \( t > 0 \), ensuring that the logarithm is defined
  • This simple rule makes it easy to handle problems involving the logarithm.
  • In our example, by differentiating \( 1 + \ln t \), the constant \( 1 \) disappears as constants have zero derivatives, leaving the term \( \frac{1}{t} \).
Combining the derivative of the logarithm with other differentiation rules like the chain rule further extends calculus tools, enabling the processing of more intricate functions.
The Power Rule for Differentiation
The power rule is one of the cornerstone techniques in calculus for finding derivatives. It is particularly convenient when dealing with polynomial terms, directly applied to powers of variables.
The rule itself is simple and practical: if you have a function \( f(x) = x^n \), where \( n \) is any real number, the derivative \( f'(x) \) is given by \( nx^{n-1} \). This applies not only for integers \( n \) but also for fractional and negative powers
  • Using the power rule saves time and simplifies differentiation steps significantly.
  • In our context of the exercise, the function \( (1 + \ln t)^{1/2} \) is an ideal candidate for the power rule.
Converting \( \sqrt{1 + \ln t} \) to \( (1 + \ln t)^{1/2} \) allows easier application of the power rule. After applying both the power and chain rules, these concepts work together to break down and solve complex calculus problems efficiently.

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