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Derivatives from a Different Perspective

Typically, the derivative of a function is defined as:

\[\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\]

This definition assumes the method of picking some horizontal distance, \(h\), and using the new x-coordinate (which is \(x + h\)) to find the new y-coordinate (which is \(f(x + h)\)), creating a secant line, and then computing the slope. When \(h \to 0\), then we can find the value/equation of the slope of the tangent line (i.e., the derivative)

  1. Copy the image below (the equation of the graph does not matter, as long as it is "curvy"), and then label \(x\), \(x + h\), \(f(x)\), \(f(x + h)\), \(h\), and the secant line.

    Now we are going to consider a different perspective, which is isntead of using a fixed horizontal distance of \(h\), we will use a fixed vertical distance of \(h\). The horizontal distance between points will be now called \(k\). Then we will still have \(h \to 0\), which we will then use a limit, along with a new equation for the derivative, to find derivatives.

  2. Make another copy of the diagram above and now label \(x\), \(x + h\), \(f(x)\), \(h\), \(f(x) + h\), and the secant line.

  3. Using algebra show that \(k = f^{-1}(f(x) + h) - x\).
  4. Now that we know \(k\), use a limit to define a new equation for the derivative of a function in terms of \(h\), \(x\), and \(f(x)\).
  5. Use your new definition of the derivative to find the derivative of the following:
    1. \(f(x) = \sqrt{x}\)
    2. \(f(x) = \frac{1}{x}\)
  6. Now consider the function \(f(x) = x^2\). Using your new definition of the derivative get to a place of evaluating the derivative up to the point where you get a \(\sqrt{x^2}\).
    1. What is \(\sqrt{x^2}\) precisely equal to?
    2. What, then becomes a difficulty when considering the derivative of \(x^2\) if we consider when \(x\) is a negative value? What is a way to resolve this issue?

Solution

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