Derivative at a Point: Limit Definition

The "Algorithm" for Finding the Derivative at a Point

Calculating the derivative from its definition, $f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$, is a methodical process. Given your background in IT, you can think of it as executing a specific algorithm.

Here are the key steps:

1. Set Up the Limit: Substitute the function $f$ and the specific point $a$ into the difference quotient formula.
2. Substitute and Expand: Replace $f(a+h)$ and $f(a)$ with their actual expressions. Expand any polynomials or distribute terms as needed. The goal is to get the numerator into a form where you can combine terms.
3. Simplify and Cancel: This is the crucial step. After simplifying the numerator, every term that does not contain $h$ should cancel out. If they don't, it's a sign of an algebraic error. Once they have canceled, you can factor an $h$ from the remaining terms in the numerator and cancel it with the $h$ in the denominator. This resolves the $\frac{0}{0}$ indeterminate form.
4. Evaluate the Limit: With the problematic $h$ in the denominator gone, you can now find the limit by substituting $h=0$ into the simplified expression. The result is the value of the derivative, $f^{\prime }(a)$.

Example 1: A Cubic Function

Let's apply this process to a cubic function. Find the derivative of $f(x)=x^3$ at $a=2$.

1. Set Up:
   $f^{\prime }(2)={\lim }_{h\to 0}\frac{f(2+h)-f(2)}{h}$

2. Substitute and Expand: We know $f(2)=2^3=8$. For $f(2+h)$, we expand the binomial $(2+h{)}^3$. Recall the binomial expansion $(x+y{)}^3=x^3+3x^{2}y+3x{y}^2+y^3$.
   $(2+h{)}^3=2^3+3(2^2)h+3(2)h^2+h^3=8+12h+6h^2+h^3$
   Substituting this into our limit:
   $f^{\prime }(2)={\lim }_{h\to 0}\frac{(8+12h+6h^2+h^3)-8}{h}$

3. Simplify and Cancel: The '8' and '-8' in the numerator cancel out.
   $f^{\prime }(2)={\lim }_{h\to 0}\frac{12h+6h^2+h^3}{h}$
   Now, we factor out $h$ from the numerator and cancel:
   $f^{\prime }(2)={\lim }_{h\to 0}\frac{h(12+6h+h^2)}{h}={\lim }_{h\to 0}(12+6h+h^2)$

4. Evaluate: Substitute $h=0$:
   $f^{\prime }(2)=12+6(0)+(0{)}^2=12$
   The derivative of $f(x)=x^3$ at $a=2$ is 12. This means the slope of the tangent line to the curve at the point $(2,8)$ is 12.

Example 2: A Radical Function

When the function involves a square root, the simplification step requires a different algebraic tool: rationalizing the numerator by multiplying by its conjugate.

The following video demonstrates this technique perfectly.

As you saw, the key steps were:

1. Setting up the limit: $f^{\prime }(4)={\lim }_{x\to 4}\frac{\sqrt{x}-\sqrt{4}}{x-4}$. (Note: This example uses the alternate $x\to a$ form, which is equivalent).
2. Multiplying the numerator and denominator by the conjugate $\sqrt{x}+2$.
3. Simplifying the numerator to $x-4$, which then cancels with the denominator.
4. Evaluating the final, simplified limit to get $1/4$.

Example 3: A Rational Function

For rational functions (fractions with polynomials), the challenge is often a "complex fraction" inside the limit. The key here is to find a common denominator to combine the fractions in the numerator into a single expression.

The next example in the same video addresses this situation.

The core algebraic steps in the video were:

1. Setting up the limit involving the complex fraction.
2. Finding the common denominator, $5(x+2)$, for the fractions in the numerator.
3. Combining these fractions into a single expression: $\frac{3-x}{5(x+2)}$.
4. Recognizing that $(3-x)=-1\cdot (x-3)$ to enable cancellation.
5. Evaluating the resulting limit to get $-1/25$.

Putting It All Together: A Practice Problem

Let's try a problem that requires combining these techniques.
Problem: Find the derivative of $f(x)=\frac{1}{\sqrt{x}}$ at $a=4$.

Let's walk through the solution using our algorithm.

1. Set Up the Limit:

   $$f^{\prime }(4)=\underset{h\to 0}{\lim }\frac{f(4+h)-f(4)}{h}=\underset{h\to 0}{\lim }\frac{\frac{1}{\sqrt{4+h}}-\frac{1}{\sqrt{4}}}{h}=\underset{h\to 0}{\lim }\frac{\frac{1}{\sqrt{4+h}}-\frac{1}{2}}{h}$$

2. Simplify the Complex Fraction (Common Denominator):
   First, we combine the two terms in the numerator. The common denominator is $2\sqrt{4+h}$.

   $$f^{\prime }(4)=\underset{h\to 0}{\lim }\frac{\frac{2-\sqrt{4+h}}{2\sqrt{4+h}}}{h}$$

   To simplify this complex fraction, we can write it as:

   $$f^{\prime }(4)=\underset{h\to 0}{\lim }\frac{2-\sqrt{4+h}}{h\cdot 2\sqrt{4+h}}$$

3. Simplify and Cancel (Rationalize):
   The numerator $2-\sqrt{4+h}$ still prevents us from substituting $h=0$. We rationalize it by multiplying by its conjugate, $2+\sqrt{4+h}$.

   $$f^{\prime }(4)=\underset{h\to 0}{\lim }\frac{2-\sqrt{4+h}}{h\cdot 2\sqrt{4+h}}\cdot \frac{2+\sqrt{4+h}}{2+\sqrt{4+h}}$$

   Multiplying the numerators gives $2^2-(\sqrt{4+h}{)}^2=4-(4+h)=-h$.

   $$f^{\prime }(4)=\underset{h\to 0}{\lim }\frac{-h}{h\cdot 2\sqrt{4+h}\cdot (2+\sqrt{4+h})}$$

   Now we can cancel the $h$ from the numerator and denominator.

   $$f^{\prime }(4)=\underset{h\to 0}{\lim }\frac{-1}{2\sqrt{4+h}\cdot (2+\sqrt{4+h})}$$

4. Evaluate the Limit:
   Finally, substitute $h=0$.

   $$f^{\prime }(4)=\frac{-1}{2\sqrt{4+0}\cdot (2+\sqrt{4+0})}=\frac{-1}{2\sqrt{4}\cdot (2+\sqrt{4})}=\frac{-1}{2(2)\cdot (2+2)}=\frac{-1}{4\cdot 4}=-\frac{1}{16}$$

The derivative of $f(x)=\frac{1}{\sqrt{x}}$ at $a=4$ is $-\frac{1}{16}$.

When the Derivative Does Not Exist

It's important to recognize that this process can fail. If the limit of the difference quotient does not exist, then the derivative at that point does not exist. This typically happens where a graph has a sharp corner, a discontinuity, or a vertical tangent line. At such points, we cannot define a single, unique tangent line.

Conclusion

In this lesson, we focused on the practical application of the limit definition to calculate derivatives. You've seen that while the definition itself is always the same, the type of function dictates the specific algebraic strategy needed to evaluate the limit.

Key Takeaways:

• Calculating a derivative from the definition is a four-step process: set up, expand, simplify/cancel, and evaluate.
• The simplification step is the most critical and requires specific algebraic tools:
  • Polynomials: Expand and combine like terms.
  • Radical Functions: Rationalize the numerator by multiplying by the conjugate.
  • Rational Functions: Simplify complex fractions by finding a common denominator.
• A derivative $f^{\prime }(a)$ fails to exist if the limit of the difference quotient does not exist, which corresponds geometrically to a "sharp corner" or other discontinuity in the graph of $f(x)$ at $x=a$.

So far, we have found the derivative one point at a time. This is powerful, but inefficient if we want to know the slope at many points. In our next lesson, we will take a major step forward by treating the point '$a$' as a variable '$x$'. This will allow us to derive a general formula, $f^{\prime }(x)$, called the derivative function, which gives the slope of the tangent line at any point on the curve.