Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 9-16, find \(f^{\prime \prime}(2)\). $$ f(t)=\frac{2}{t} $$

Step-by-Step Solution

Answer

The second derivative at \( t = 2 \) is \( \frac{1}{2} \).

Step 1: Find the First Derivative

The function given is \( f(t) = \frac{2}{t} \). We need to find the first derivative \( f'(t) \). Recall that the derivative of \( t^{-1} \) is \( -t^{-2} \). Therefore, rewrite \( f(t) \) as \( 2t^{-1} \) and find the derivative:\[ f'(t) = \frac{d}{dt}(2t^{-1}) = 2(-t^{-2}) = -\frac{2}{t^2} \]

Step 2: Find the Second Derivative

Now, we find the second derivative \( f''(t) \) by differentiating \( f'(t) = -\frac{2}{t^2} \). We apply the power rule again, where the derivative of \( t^{-2} \) is \( -2t^{-3} \):\[ f''(t) = \frac{d}{dt}(-2t^{-2}) = -2(-2t^{-3}) = \frac{4}{t^3} \]

Step 3: Evaluate the Second Derivative at \( t = 2 \)

Now that we have the second derivative, we evaluate \( f''(t) \) at \( t = 2 \):\[ f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} \]

Step 4: Conclusion

We have found that the value of the second derivative of the function \( f(t) = \frac{2}{t} \) at \( t = 2 \) is \( \frac{1}{2} \).

Key Concepts

Derivatives

Understanding derivatives is fundamental to learning calculus. In simple terms, a derivative represents the rate of change of a function as its input changes. Imagine you are driving a car, and the speedometer shows your speed at any given moment. This speed is a real-world example of a derivative, showing how your distance changes over time.
To find a derivative, we apply differentiation rules, which are systematic steps that help us compute derivatives for various types of functions. In our original problem, the function given is \(f(t) = \frac{2}{t}\). To differentiate it, we first rewrite it in a more manageable form \(f(t) = 2t^{-1}\). From here, we apply the basic principles of differentiation to find both the first and the second derivatives.

Power Rule

The power rule is one of the most important tools for finding derivatives quickly and easily. The rule says that if you have a function \(x^n\), its derivative is \(nx^{n-1}\). This rule is particularly useful for various forms of algebraic expressions involving powers of variables.

In our function \(f(t) = 2t^{-1}\), the power rule allows us to differentiate efficiently. By applying the power rule to get the first derivative, we derive \(f'(t) = -\frac{2}{t^2}\) by treating \(t^{-1}\) as our power term and multiplying by the constant 2, while reducing the power by one. This process simplifies the function and prepares it for further differentiation.

Second Derivative

The second derivative provides us with even deeper insights into a function's behavior. If the first derivative shows the rate of change, the second derivative tells us about the rate of that rate of change. This is like how acceleration tells you how your speed is changing in a vehicle.

It is calculated by differentiating the first derivative of a function. In our exercise, the first derivative \(f'(t) = -\frac{2}{t^2}\) is further differentiated using the power rule again. This gives us the second derivative \(f''(t) = \frac{4}{t^3}\). By evaluating this derivative at a specific point \(t = 2\), we find \(f''(2) = \frac{1}{2}\), indicating a particular rate at which the rate of change is happening at that point. Thus, second derivatives hold key information about the concavity and potential inflection points of the function graph.