Functions
Limits and Derivatives
Applications of Differentiation
Integration

The Derivative as a Function

The Derivative as a Function

Recall the definition of the (\(h\)-form of the) derivative at a given \(x\)-value.

Now suppose that we want to use this definition of the derivative to find the derivative of the function \(f(x)=x^2\) at several different \(x\)-values, say \(x=1,2,3\) and \(4\). The process for finding the derivative in each of these four cases is the same; as is the computations required to compute the final answer. This is exhibited in the table below.

table of derivative results for the function x^2 at the x-values x=1,2,3,4

Since the process of computing this derivative is basically the same, with the only thing that changes being the \(x\)-values, we could surmise that the process would be the same for any other \(x\)-value we choose. So, since the process is the same regardless of the \(x\)-value, why don’t we simply use a “placeholder symbol” for our \(x\)-value and carry out the process as usual. Let’s call this placeholder (for lack of more appropriate options) “\(x\).” Then the derivative at \(x\) is computed as:

steps for the derivative of x^2 which eventually gives you a result of 2x

Since \(x\) was a placeholder for any \(x\)-value we wanted to find the derivative at, and given that everything above is equal, we essentially have a formula that will tell us the derivative of \(f(x)=x^2\) no matter what \(x\) is. You can confirm this by plugging \(x=1,2,3,4\) into the formula \(f'(x)=2x\) and compare with the derivatives in the table we built earlier to see that this formula works.

In general, this idea allows us to build a formula that gives us the slope (or instantaneous rate of change) of a function at any \(x\)-value we like.

Try some of the examples below for finding the derivative of various functions!

Find the derivative of the following functions

$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(c)}{h}$$

  • Compute \(f(x+h)\) first by plugging the expression \((x+h)\) into \(f\)
  • Compute and simplify \(f(x+h)-f(x)\) using the answer to the previous step
  • Compute and simplify \(\frac{f(x+h)-f(x)}{h}\) using the answer to the previous step
  • Take the limit \(\lim_{h\rightarrow 0}\) of the answer to the above.

\(f'(x)=3\)

$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(c)}{h}$$

  • Compute \(f(x+h)\) first by plugging the expression \((x+h)\) into \(f\)
  • Compute and simplify \(f(x+h)-f(x)\) using the answer to the previous step
  • Compute and simplify \(\frac{f(x+h)-f(x)}{h}\) using the answer to the previous step
  • Take the limit \(\lim_{h\rightarrow 0}\) of the answer to the above.

\(f'(x)=5\)

$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(c)}{h}$$

  • Compute \(f(x+h)\) first by plugging the expression \((x+h)\) into \(f\)
  • Compute and simplify \(f(x+h)-f(x)\) using the answer to the previous step
  • Compute and simplify \(\frac{f(x+h)-f(x)}{h}\) using the answer to the previous step
  • Take the limit \(\lim_{h\rightarrow 0}\) of the answer to the above.

\(f'(x)=m\). Keep in mind that the derivative is a “slope-producing formula,” and that for a line in the form \(y=mx+b\) the slope is \(m\), which matches our derivative. So derivatives do what they’re supposed to.

$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(c)}{h}$$

  • Compute \(f(x+h)\) first by plugging the expression \((x+h)\) into \(f\)
  • Compute and simplify \(f(x+h)-f(x)\) using the answer to the previous step
  • Compute and simplify \(\frac{f(x+h)-f(x)}{h}\) using the answer to the previous step
  • Take the limit \(\lim_{h\rightarrow 0}\) of the answer to the above.

\(f'(x)=6x\), and so

\(f'(1)=6\), \(f'(2)=12\)

$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(c)}{h}$$

  • Compute \(f(x+h)\) first by plugging the expression \((x+h)\) into \(f\)
  • Compute and simplify \(f(x+h)-f(x)\) using the answer to the previous step
  • Compute and simplify \(\frac{f(x+h)-f(x)}{h}\) using the answer to the previous step
  • Take the limit \(\lim_{h\rightarrow 0}\) of the answer to the above.

\(g'(x)=-4x\), and so

\(g'(2)=-8\), \(g'(-1)=4\)

$$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(c)}{h}$$

  • Compute \(f(x+h)\) first by plugging the expression \((x+h)\) into \(f\)
  • Compute and simplify \(f(x+h)-f(x)\) using the answer to the previous step
  • Compute and simplify \(\frac{f(x+h)-f(x)}{h}\) using the answer to the previous step
  • Take the limit \(\lim_{h\rightarrow 0}\) of the answer to the above.

\(g'(x)=3x^2\), and so

\(g'(1)=3\), \(g'(-1)=3\)

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