Problem 43 - Applied Calculus

Question

If \(\frac{d}{d t}(t f(t))=1+f(t)\), what is \(f^{\prime}(t) ?\)

Step-by-Step Solution

Answer

\(f'(t) = \frac{1}{t}\).

Step 1: Differentiate the Product

Begin by applying the product rule to differentiate the expression \(t f(t)\). The product rule states that if you have two functions \(u(t)\) and \(v(t)\), the derivative is \(u'(t)v(t) + u(t)v'(t)\). Here, let \(u(t) = t\) and \(v(t) = f(t)\). Thus, \(\frac{d}{dt}(t f(t)) = 1 \cdot f(t) + t \cdot f'(t)\).

Step 2: Set the Derivative Equal to Given Expression

From the problem, we have that \(\frac{d}{dt}(t f(t)) = 1 + f(t)\). Substitute the expression from Step 1: \(f(t) + t f'(t) = 1 + f(t)\).

Step 3: Simplify the Equation

Subtract \(f(t)\) from both sides to isolate the term involving \(f'(t)\): \(t f'(t) = 1\).

Step 4: Solve for \(f'(t)\)

To solve for \(f'(t)\), divide both sides by \(t\): \(f'(t) = \frac{1}{t}\).

Key Concepts

Product Rule

In calculus, the product rule is a key technique used to differentiate expressions that are the product of two functions. This can be particularly useful when you’re dealing with complex equations involving multiple functions multiplied together.
Here’s how the product rule works:

• Let’s assume we have two functions, say, \( u(t) \) and \( v(t) \). The product rule gives us a formula to find the derivative of their product \( u(t)v(t) \).
• According to the product rule, the derivative of \( u(t) \) times \( v(t) \) is given by:

\[\frac{d}{dt}[u(t) \cdot v(t)] = u'(t) \cdot v(t) + u(t) \cdot v'(t)\]This rule is incredibly handy when one or both functions are not simple or are difficult to differentiate directly.
In our original exercise, we applied the product rule to the expression \( t f(t) \) by letting \( u(t) = t \) and \( v(t) = f(t) \). Following the product rule, the derivative is \( 1 \cdot f(t) + t \cdot f'(t) \). This initial step is crucial as it sets up the solution by breaking down the problem into more manageable parts.

Differentiation

Differentiation is one of the fundamental concepts of calculus. It allows us to find the *rate of change* or the *slope* of a function at any given point. This is often synonymous with calculating the derivative.
Differentiation is useful in many fields, from physics to economics, enabling us to model real-world behaviors and dynamics.

• When you differentiate a constant, the result is zero because constants do not change.
• The power rule is common in differentiation. For example, the derivative of \( t^n \) is \( n \cdot t^{n-1} \).

Differentiation can be applied to polynomials, exponential functions, trigonometric functions, and more.
In the exercise, differentiating the expression \( t f(t) \) by using the product rule resulted in the derivative expression \( f(t) + t f'(t) \). This process highlights how differentiation can be managed even when functions are multiplied together.

Derivative

The derivative of a function is the mathematical tool that tells us how the function changes at any given point. It's the foundation of differential calculus and provides vital insights into the behavior of functions.
When we compute a derivative, we're essentially finding the slope of the tangent line to the function’s graph at a particular point. This gives us information about the function’s increasing or decreasing trends.

• The notation \( f'(t) \) represents the derivative of the function \( f(t) \).
• Derivatives can tell us about the concavity and inflection points of a function.

The exercise asked us to find \( f'(t) \), the derivative of the function \( f(t) \). By using known differentiation techniques like the product rule, we simplified the given expression and isolated \( f'(t) \) as \( \frac{1}{t} \). This result shows a clear summary of how the function \( f(t) \) changes with respect to \( t \).