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In calculus, the derivative of a function measures how the function's output changes as its input changes. Imagine you're driving a car; the derivative is like the speedometer telling you how fast you're going at every moment. That's its significance in a broader context. When you take the derivative of a function, you often want to understand how the function behaves and predict its future values.
To differentiate a function means to find this rate of change. For our function, \(f(x) = \frac{1}{x}\), we found the derivative to be \(f'(x) = -\frac{1}{x^2}\). This means that as x changes, the rate at which \(f(x)\) changes is given by \(-\frac{1}{x^2}\). This gives us insight into how drastic or mild these changes are, depending on the value of x.

• The derivative tells us about the slope of the tangent line to the graph of \(f(x)\) at any point \(x\).
• If \(f'(x)\) is positive, \(f(x)\) is increasing; if it's negative, \(f(x)\) is decreasing.
• The magnitude of \(f'(x)\) tells us how steep the curve is at a particular point.

The power rule is a fundamental tool for taking derivatives that simplifies the process greatly. It is essential for students to understand this rule, as it applies in many situations. The power rule, which states \(\frac{d}{dx}[x^n] = n \cdot x^{n-1}\), makes finding the derivative of polynomial functions straightforward.
For the function \(f(x) = \frac{1}{x}\), we recognize this as being \(x^{-1}\), which fits our power rule perfectly with \(n = -1\). By applying the power rule:

• Differentiate \(x^{-1}\) to get \(-1 \cdot x^{-2}\).
• The derivative simplifies to \(-\frac{1}{x^2}\), as seen in the previous section.

This approach is effective because it reduces what might seem complex into a series of manageable steps using a systematic rule. Understanding and using the power rule can save a lot of time and effort when faced with functions similar to \(x^{-1}\). It's a valuable shortcut that becomes second nature with practice.

Once you have the derivative, evaluating it at certain points can yield critical information about the original function and its behavior. Let's break down how to do this with examples from the solution:

• For \(f'(0)\), we would encounter a division by zero in \(-\frac{1}{x^2}\), meaning \(f'(0)\) is undefined. It indicates a point where the function might have a discontinuity or an asymptote.
• Evaluating \(f'(2)\), you substitute 2 for \(x\) in \(-\frac{1}{x^2}\). This gives \(f'(2) = -\frac{1}{4}\). This indicates that at \(x = 2\), the function is decreasing with a slope of \(-\frac{1}{4}\).
• For \(f'(-1)\), substitute \(-1\) for \(x\) to get \(f'(-1) = -1\). This means at \(x = -1\), the function is decreasing more steeply with a slope of \(-1\).

By evaluating the derivative at specific points, you gain a clear picture of how the original function behaves around those values. It provides a snapshot of the function's incline or decline, which is essential for sketching curves and analyzing trends in calculus.