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Magazine Chapter 2: Problem 6
Differentiate the function. $$ g(t)=t \ln 2 t $$
Short Answer
Expert verified
The derivative of \(g(t) = t \ln(2t)\) is \(g'(t) = \ln(2t) + 1\).
Step by step solution
01
Identify the two functions in the product
As mentioned earlier, our function \(g(t)=t \ln(2t)\) has two component functions: \(f(t) = t\) and \(g(t) = \ln(2t)\).
02
Differentiate the two functions with respect to t
Now, we will differentiate both functions. The derivative of \(f(t) = t\) is simply:
\[f'(t) = 1\]
For \(g(t) = \ln(2t)\), we need to apply the chain rule since it's a composition of functions. The chain rule states that if we have a function, \(g(t) = h(k(t))\), then the derivative of \(g(t)\) is given by:
\[g'(t) = h'(k(t)) \cdot k'(t)\]
Here, we have \(h(t) = \ln t\) and \(k(t) = 2t\), so we get:
\(h'(t) = \frac{1}{t}\) and \(k'(t) = 2\).
Now, applying the chain rule:
\[g'(t) = h'(k(t)) \cdot k'(t) = \frac{1}{2t} \cdot 2 = \frac{1}{t}\]
03
Apply the product rule
Finally, we will apply the product rule to find the derivative of our original function:
\[g'(t) = f'(t) \cdot g(t) + f(t) \cdot g'(t) = (1) \cdot (\ln(2t)) + (t) \cdot \left(\frac{1}{t}\right)\]
Simplifying this expression, we get:
\[g'(t) = \ln(2t) + 1\]
Thus, the derivative of the function \(g(t) = t \ln(2t)\) is:
\[g'(t) = \ln(2t) + 1\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
Understanding the product rule is essential when dealing with derivatives of functions that are the product of two or more functions. When you have a function that can be expressed as the product of two simpler functions, say, function \( f(t) \) and function \( g(t) \), the product rule allows you to find the derivative of this composite function.
The product rule states that if you have two functions \( f(t) \) and \( g(t) \), the derivative of their product \( f(t) \times g(t) \) is given by:
\[ f'(t) \times g(t) + f(t) \times g'(t) \]
This rule is applied by differentiating each function independently (\( f'(t) \) and \( g'(t) \)) and then using these derivatives to calculate the derivative of the product. Thus, it breaks down a potentially complex differentiation problem into simpler parts.
The product rule states that if you have two functions \( f(t) \) and \( g(t) \), the derivative of their product \( f(t) \times g(t) \) is given by:
\[ f'(t) \times g(t) + f(t) \times g'(t) \]
This rule is applied by differentiating each function independently (\( f'(t) \) and \( g'(t) \)) and then using these derivatives to calculate the derivative of the product. Thus, it breaks down a potentially complex differentiation problem into simpler parts.
Chain Rule
The chain rule is another fundamental tool in calculus for differentiating composite functions. When a function \( g(t) \) is composed of two functions, such that \( g(t) = h(k(t)) \), the chain rule helps you find the derivative by taking into account the relationship between the two functions.
The chain rule formula is:
\[ g'(t) = h'(k(t)) \times k'(t) \]
You first differentiate the outer function \( h(t) \) with respect to its inner function \( k(t) \) to get \( h'(k(t)) \), then multiply by the derivative of the inner function \( k'(t) \). This process is tAlso crucial for functions with multiple layers of composition, as it can be applied repeatedly to peel back each layer until you reach the derivative of the innermost function.
The chain rule formula is:
\[ g'(t) = h'(k(t)) \times k'(t) \]
You first differentiate the outer function \( h(t) \) with respect to its inner function \( k(t) \) to get \( h'(k(t)) \), then multiply by the derivative of the inner function \( k'(t) \). This process is tAlso crucial for functions with multiple layers of composition, as it can be applied repeatedly to peel back each layer until you reach the derivative of the innermost function.
Derivatives
Derivatives represent the rate at which a function is changing at any given point and are a foundational concept in calculus. The derivative of a function at a certain point gives you the slope of the tangent line to the function's graph at that point, which indicates how the function value is changing with respect to a change in its input value.
Simply put, if you have a function \( y = f(x) \), the derivative of this function is often written as \( f'(x) \) or \( \frac{dy}{dx} \), and it measures how \( y \) changes as \( x \) changes. Derivatives can be used to find maximums, minimums, and inflection points of a function, making them a powerful tool in various scientific and engineering fields.
Simply put, if you have a function \( y = f(x) \), the derivative of this function is often written as \( f'(x) \) or \( \frac{dy}{dx} \), and it measures how \( y \) changes as \( x \) changes. Derivatives can be used to find maximums, minimums, and inflection points of a function, making them a powerful tool in various scientific and engineering fields.
Logarithmic Differentiation
Logarithmic differentiation is a method that can simplify the differentiation of complex functions, especially those involving products, quotients, and powers of functions. It is particularly useful when the function is given in a form that is hard to differentiate using standard rules.
In logarithmic differentiation, you take the natural logarithm of both sides of an equation \( y = f(x) \) and then differentiate using the properties of logarithms to break down the product into sums, the quotient into differences, and the powers into products. This can greatly simplify the differentiation process. After differentiating, you solve for \( y' \), the derivative of the original function. It's important to remember that since you've taken the logarithm of the function, you'll need to use the chain rule to complete the differentiation process.
In logarithmic differentiation, you take the natural logarithm of both sides of an equation \( y = f(x) \) and then differentiate using the properties of logarithms to break down the product into sums, the quotient into differences, and the powers into products. This can greatly simplify the differentiation process. After differentiating, you solve for \( y' \), the derivative of the original function. It's important to remember that since you've taken the logarithm of the function, you'll need to use the chain rule to complete the differentiation process.
- Applicable when differentiating variables in an exponent
- Utilizes the logarithm properties to simplify complex functions before differentiation
- Often combined with the chain rule for composite functions involving logarithms
