Calculus I Lesson 7: Derivative Rules

  1. Warm up
  2. Notations
  3. Product Rule
    1. Exercises
  4. Power Rule
    1. Exercises
  5. Differentiability and Continuity
    1. Differentiable, Continuous, both, neither
    2. x = 0
    3. x = 1
    4. x = 1
    5. x = 2
    6. Graph
  6. Upcoming

Warm up

Consider the following function:

\[f(x) = \begin{cases}kx^2 + 2 &\mbox{if } x < 1 \\ 4x - 1 &\mbox{if } x \geq 1 \end{cases}\]

A student is asked to find $k$ such that $f$ is differentiable at $x = 1$. The student does the following:

  1. When $x < 1$, using the sum and constant multiple rules, $f^\prime(x) = 2kx$.
  2. When $x \geq 1$, since $4x - 1$ is a line with slop e1, $f^\prime(x) = 4$.
  3. Setting these equal to each other when $x = 1$ and solving, we get that $k = 2$.

Critique the student’s response.

Notations

There are a few ways in which we might refer to taking a derivative. They all mean the same thing

The $\dfrac{dy}{dx}$ notation is meant to evoke the “average rate of change” $\dfrac{\Delta y}{\Delta x}$. The idea is, as $\Delta x \rightarrow 0$, $\dfrac{\Delta y}{\Delta x} \rightarrow \dfrac{dy}{dx}$.

Product Rule

We have seen a few examples of derivatives already, and also learned the sum and constant multiple rules. Today we will learn the product rule and power rules. Let’s review an important example first: the derivative of $x^2$ is $2x$. We figured this out by looking at In this video, I take a geometric look at this derivative.

$x^2$ represents the area of a square whose side length is $x$. The derivative represents the rate at which the derivative changes in terms of how fast $x$ is changing. So how does this change? Let $\Delta x$ represent a small change in the side length. Then the change in the area is $2x\Delta x + (\Delta x)^2$. The rate of change of the area, with respect to changing $x$, is $\dfrac{2x \Delta x + (\Delta x)^2}{\Delta x}$, which, as $\Delta x \rightarrow 0$, appraoches $2x$.

Similarly, we can think of a product of two functions $f(x) \cdot g(x)$ as representing the area of a rectangle whose side lengths are $f(x)$ and $g(x)$. What happens as we change $x$ by a small amount, say $\Delta x$? The area changes by $(\Delta f)g + (\Delta g)f + \Delta f \Delta g$. The rate of change of the area, with respect to changing $x$, is $\dfrac{(\Delta f)g + (\Delta g)f + \Delta f \Delta g}{\Delta x}$, which is $\dfrac{\Delta f}{\Delta x}g + \dfrac{\Delta g}{\Delta x}f + \dfrac{\Delta f}{\Delta x}\Delta g$ As $x \rightarrow 0$, this approaches $f^\prime(x)g(x) + g^\prime(x)f(x)$. The last term goes away since, as $\Delta x \rightarrow 0$, $\Delta g \rightarrow 0$ also. This is the product rule for derivatives!

Product rule: Let $f$ and $g$ be functions that are differentiable at $x$. Then $(f(x)g(x))^\prime = f^\prime(x)g(x) + g^\prime(x) f(x)$.

Exercises

  1. Use the sum rules from last time to find the derivative of $x^2 + x + 1$.
  2. Find the derivative of $x - 1$.
  3. Find the derivative of $(x-1)(x^2 + x + 1)$ using the product rule.
Solution:
  1. $2x + 1$
  2. 1
  3. Using the product rule: $1(x^2 + x + 1) + (x-1)(2x + 1)$. This simplifies to $x^2 + x + 1 + 2x^2 + x - 2x - 1$, which is $3x^2$

The next set of exercises is meant to help you discover the power rule.

  1. Use the fact that $(x^2)^\prime = 2x$ and $x^\prime = 1$ to find $(x^3)^\prime$. (Hint: $x^3 = x^2 \cdot x$.)
  2. Use your answer to part 1, and the fact that $x^4 = x^3 \cdot x$ to find the derivative of $x^4$.
  3. Use your answer to part 2 to find the derivative of $x^5$.

Can you find a pattern?

Power Rule

Hopefully the exercises in the previous part helped you figure out the power rule.

Power rule (part 1): Let $n > 0$ be a positive integer. Then the derivative of $x^n$ is $nx^{n-1}$.

This rule will actually generalize to other powers of $x$: like $x^{\frac{1}{2}}$, or $x^\pi$, but we won’t study those just yet.

Combining the power rule with the other rules we learned so far helps us compute derivatives of all polynomials. For example, let $f(x) = 2x^3 - 4x^2 + 3x - 1$. To find the derivative, we find the derivative of each individual term, and add/subtract them (using the sum rule). So we need to find the derivatives of the following:

The derivative of $x^3$ is $3x^2$, by the power rule. But by the constant multiple rule, that means the derivative of $2x^3$ is $2 \times 3x^2$, which is $6x^2$. Similarly, the derivative of $4x^2$ is $4 \times 2x$, which is $8x$. The derivative of $3x$ is $3$, and the derivative of $-1$ is $0$, since it’s a constant function.

That means the derivative of $2x^3 - 4x^2 + 3x - 1$ is $6x^2 - 8x + 3$.

Moral of the story: Given a polynomial, the derivative can be found by using the power rule on each individual term, multiplying by the coefficient of that term, and adding the terms together.

Exercises

Use the power rule to find the derivative of the following:

  1. $f(x) = 3x^4 + 4x^3 + 5x^2 + 6x + 7$
  2. $g(x) = x^3 + x^2 + x + 1$

Use the power rules and product rules, together, to find the derivatives of the following:

  1. $h(x) = (x^3 - 1)(3x^2 - 2x + 1)$
  2. $j(x) = (x+1)^2$, using the fact that $(x+1)^2$ is $(x+1)(x+1)$.
  3. $k(x) = (4x - 3)^2$.
  4. $F(x) = (3x - 1)^3$. Here you would need to first use the product rule to find the derivative of $(3x - 1)^2$, and then again use the product rule to find the derivative of $(3x - 1)^2 \cdot (3x - 1)$.

Differentiability and Continuity

Let’s look at some graphs of discontinuous functions. First, with a hole:

What do you notice about the secant lines to the right and left of 0?

Now let’s look at a graph with a jump:

Based on the graphs we’ve seen, which of the following statements are true?

Now let’s look at a graph of a continuous function.

Based on the previous graphs, what conclusions can we draw? Explain your choices.

Theorem: Suppose $f(x)$ is differentiable at $x = a$. Then $f(x)$ is continuous at $x = a$.

Proof: The limit as $x \rightarrow a$ of $\dfrac{f(x)-f(a)}{x - a}$ must exist. Suppose it’s equal to $L$. Then, as $x \rightarrow a$, $f(x) - f(a) \rightarrow (x - a)L$, which approaches $0$. That means, as $x \rightarrow a$, $f(x) \rightarrow f(a)$.

Differentiable, Continuous, both, neither

\[f(x) = \begin{cases}x^2 - 1 &\mbox{if } x < 1 \\ x - 1 &\mbox{if } x \geq 1 \end{cases}\]

Is $f(x)$ differentiable, continuous, both, or neither at the following points:

x = 0

$x^2 - 1$ is differentiable at $x = 0$. Therefore it is also continuous.

Answer: Both

x = 1

Need to check left and right hand limits of

\[\lim_{x \rightarrow 1} \frac{f(x) - f(1)}{x - 1}\]

From the left: $\dfrac{(x^2 - 1) - 0}{x - 1} \rightarrow x + 1 \rightarrow 2$.

From the right: $\dfrac{x -1 - 0}{x - 1} \rightarrow 1$. Not differentiable

x = 1

Now we check if $f(x)$ is continuous at $x = 1$:

So $f$ is continuous at $x = 1$.

x = 2

$x - 1$ is differentiable, its derivative is $1$ near $x = 2$. Differentiable functions are continuous, so both.

Graph

Upcoming