Problem 46 - University Calculus: Early Transcendentals

Question

Find the first and second derivatives of the functions. $$s=\frac{t^{2}+5 t-1}{t^{2}}$$

Step-by-Step Solution

Answer

First derivative: \( s' = -5t^{-2} + 2t^{-3} \), Second derivative: \( s'' = 10t^{-3} - 6t^{-4} \).

Step 1: Convert to Simplified Form

Before finding the derivative, express the function in a more manageable form by dividing each term in the numerator by the denominator. The function is \( s = \frac{t^2}{t^2} + \frac{5t}{t^2} - \frac{1}{t^2} \). Simplify it to: \( s = 1 + \frac{5}{t} - \frac{1}{t^2} \), which becomes \( s = 1 + 5t^{-1} - t^{-2} \).

Step 2: Find the First Derivative

Differentiate the function with respect to \( t \). Apply the power rule, which states that \( \frac{d}{dt} [t^n] = nt^{n-1} \).\[ s' = \frac{d}{dt} [1 + 5t^{-1} - t^{-2}] \]Calculate each term separately:- The derivative of \( 1 \) is \( 0 \). - The derivative of \( 5t^{-1} \) is \( -5t^{-2} \). - The derivative of \( -t^{-2} \) is \( 2t^{-3} \).Thus, the first derivative is:\( s' = -5t^{-2} + 2t^{-3} \).

Step 3: Find the Second Derivative

Differentiate the first derivative with respect to \( t \) again.Given \( s' = -5t^{-2} + 2t^{-3} \), apply the power rule again:Calculate each term separately:- The derivative of \( -5t^{-2} \) is \( 10t^{-3} \). - The derivative of \( 2t^{-3} \) is \( -6t^{-4} \).Thus, the second derivative is:\( s'' = 10t^{-3} - 6t^{-4} \).

Key Concepts

Power Rule

In calculus, the power rule is a fundamental differentiation technique. It is used to find the derivative of functions that are expressed as powers of a variable. Specifically, for a function of the form \(t^n\), where \(n\) is any real number, the power rule allows us to determine its derivative efficiently.

To apply the power rule, follow this simple formula:

• Formula: \( \frac{d}{dt} [t^n] = nt^{n-1} \)
• Multiply the coefficient \(n\) by the variable base \(t\), and subtract one from the exponent \(n\).

For example, if we have a term like \(t^{-2}\), applying the power rule gives us \(-2t^{-3}\). The power rule is especially useful in handling functions with negative or fractional exponents, enabling swift computation of derivatives. Remember, it's applicable to any real exponent, making it a versatile tool for differentiation.

First Derivative

The first derivative of a function provides the rate at which the function's value changes with respect to a change in the input variable. Essentially, it tells us the slope of the function at any given point.

For the function given by \(s = 1 + 5t^{-1} - t^{-2}\), finding the first derivative involves applying the power rule to each term separately:

• The constant \(1\) has a derivative of zero, as constants don't change.
• The derivative of \(5t^{-1}\) is \(-5t^{-2}\). This is because we multiply \(-1\) by 5, resulting in \(-5\) and reduce the exponent by one.
• The derivative of \(-t^{-2}\) is \(2t^{-3}\), calculated by multiplying \(-2\) by the coefficient 1, resulting in 2, while reducing the exponent by one.

Combining these, we get the first derivative: \(s' = -5t^{-2} + 2t^{-3}\). This expression shows how the function \(s\) behaves as \(t\) changes.

Second Derivative

The second derivative is the derivative of the first derivative. It provides information about the concavity of the original function. In simple terms, it tells us how the slope itself is changing.

Using the first derivative \(s' = -5t^{-2} + 2t^{-3}\), we find the second derivative by once again applying the power rule individually to each term:

• For \(-5t^{-2}\), the derivative is \(10t^{-3}\). Here, multiplying \(-2\) by \(-5\) gives us \(10\), and the exponent reduces by one.
• For \(2t^{-3}\), the derivative is \(-6t^{-4}\). By multiplying \(-3\) by 2, we get \(-6\), and again, reduce the exponent by one.

Thus, the second derivative is \(s'' = 10t^{-3} - 6t^{-4}\). This output explains how fast the rate of change of the slope (given by the first derivative) is changing with \(t\). Alternately, it sheds light on the curvature of the original function \(s\).

Simplified Form

Simplification of algebraic expressions is a vital step before applying differentiation rules. It transforms complex functions into manageable forms, which makes the calculus process straightforward.

In the problem, we started with the function \(s=\frac{t^{2}+5 t-1}{t^{2}}\). Simplifying involves distributing the denominator across each term in the numerator. This resulted in:

• \(\frac{t^2}{t^2} = 1\)
• \(\frac{5t}{t^2} = 5t^{-1}\)
• \(\frac{-1}{t^2}= -t^{-2}\)

Thus, the simplified form of the function is \(s = 1 + 5t^{-1} - t^{-2}\). This step is critical as it transforms the quotient into a sum of simpler power forms, making the application of the power rule more direct and less error-prone. Always aim for this simplified form when preparing functions for differentiation.