CSCA MATHEMATICS / MODULE 02 / LESSON 02.11

Introduction to Derivatives

From average rates of change and the difference quotient to differentiation rules, polynomials, and instantaneous velocity.

Foundational calculusWorked examplesNo standalone quiz

The central idea

\[f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]

The derivative is the limiting average rate of change at a particular input, when that finite two-sided limit exists.

02.11.1 Introduction to Derivatives

A derivative measures the instantaneous rate of change of a function with respect to its input. Another way to understand it is as the slope of a function’s graph at a particular point, provided the derivative exists.

Consider:

\[f(x)=2x+3\]

This is a linear function. Its graph is a straight line, and its slope is 2. Whenever \(x\) increases by 1, the value of \(f(x)\) increases by 2.

For example:

\[f(1)=2(1)+3=5\]
\[f(2)=2(2)+3=7\]
\[f(3)=2(3)+3=9\]

The output increases at a constant rate. Therefore, the derivative of this function is:

\[f'(x)=2\]

Now consider:

\[g(x)=x^2\]

This function does not change at a constant rate. Its graph is curved, and its slope depends on the input. The derivative is:

\[g'(x)=2x\]

At \(x=1\):

\[g'(1)=2\]

At \(x=3\):

\[g'(3)=6\]

Thus, the instantaneous rate of change is different at different points.

Why do we need derivatives?

Derivatives are used to study quantities that change continuously.

Examples include:

  • The instantaneous velocity of a moving object.
  • The rate at which a population grows.
  • How quickly a cost function increases.
  • The slope of a curved graph at a point.
  • How a function’s rate of increase or decrease varies.

In this lesson, the focus is on understanding and calculating derivatives, rather than advanced applications.

Important distinction

For a linear function:

\[f(x)=mx+b\]

the derivative is the constant \(m\). For a nonlinear function, the derivative generally depends on \(x\).

02.11.2 Average Rate of Change

Before studying instantaneous change, we must understand average rate of change. Suppose a function changes from \(f(a)\) to \(f(b)\) as its input changes from \(a\) to \(b\). The change in output is:

\[\Delta y=f(b)-f(a)\]

The change in input is:

\[\Delta x=b-a\]

Therefore, for \(a\neq b\):

\[\text{Average rate of change} =\frac{f(b)-f(a)}{b-a}\]

This is the slope of the straight line connecting the two corresponding points on the graph. That straight line is called a secant line.

Example 1: Average rate of change of a linear function

Given:

\[f(x)=3x+2\]

Find the average rate of change from \(x=1\) to \(x=5\).

Step 1. Find the first output.

\[f(1)=3(1)+2=5\]

Step 2. Find the second output.

\[f(5)=3(5)+2=17\]

Step 3. Apply the formula.

\[\frac{f(5)-f(1)}{5-1} =\frac{17-5}{4}\]
\[=\frac{12}{4}=3\]

Final answer: 3.

Because the function is linear, its rate of change is constant across the interval.

Example 2: Average rate of change of a quadratic function

Given:

\[f(x)=x^2\]

Find the average rate of change from \(x=1\) to \(x=3\).

Step 1. Evaluate at \(x=1\).

\[f(1)=1^2=1\]

Step 2. Evaluate at \(x=3\).

\[f(3)=3^2=9\]

Step 3. Calculate the average rate.

\[\frac{f(3)-f(1)}{3-1} =\frac{9-1}{2}\]
\[=\frac82=4\]

Final answer: 4.

The function’s average rate of change over the interval \([1,3]\) is 4. This does not mean the instantaneous rate equals 4 at every point in that interval.

Example 3: Comparing two intervals

Consider again:

\[f(x)=x^2\]

Calculate the average rate of change on two intervals.

From \(x=1\) to \(x=2\):

\[\frac{f(2)-f(1)}{2-1} =\frac{4-1}{1}=3\]

From \(x=2\) to \(x=3\):

\[\frac{f(3)-f(2)}{3-2} =\frac{9-4}{1}=5\]

The average rates are different. This shows that the quadratic function becomes steeper as \(x\) increases over these positive intervals.

Average Rate of Change of f(x) = x²

Function Interval Output change Average rate
x² [0, 1] 1 – 0 = 1 1
x² [1, 2] 4 – 1 = 3 3
x² [2, 3] 9 – 4 = 5 5
x² [3, 4] 16 – 9 = 7 7

02.11.3 Instantaneous Rate of Change

The instantaneous rate of change describes how quickly a function is changing at one particular input. For example, average speed measures the total distance traveled divided by total elapsed time. Instantaneous velocity describes the rate of change of position at a particular moment.

Suppose the position of an object is:

\[s(t)=t^2\]

where \(s\) is in meters and \(t\) is in seconds.

Between \(t=2\) and \(t=3\):

\[s(2)=4,\qquad s(3)=9\]

The average velocity is:

\[\frac{9-4}{3-2}=5\text{ m/s}\]

But what is the velocity at exactly \(t=2\)?

To answer this, we need the instantaneous rate of change. For the position function:

\[s(t)=t^2\]

the derivative is:

\[s'(t)=2t\]

Therefore:

\[s'(2)=2(2)=4\]

The instantaneous velocity at \(t=2\) is:

\[4\text{ m/s}\]

Notice the difference:

  • Average velocity over \([2,3]\): 5 m/s.
  • Instantaneous velocity at \(t=2\): 4 m/s.
  • Both values are correct, but they describe different quantities.

Average vs. Instantaneous Rate of Change

Feature Average rate Instantaneous rate
Describes Change over an interval Change at a point
Requires Two distinct inputs A limiting process
Graphical meaning Secant-line slope Tangent-line slope
Notation Average rate f'(a)

02.11.4 The Intuitive Concept of a Limit

The concept of a limit is essential for understanding derivatives. A limit describes the value that an expression approaches as its input approaches a particular number. For example, consider:

\[f(x)=x+2\]

As \(x\) approaches 3, the output approaches 5.

We write:

\[\lim_{x\to3}(x+2)=5\]

The symbol:

\[\lim_{x\to3}\]

is read as the limit as x approaches 3.

Example 4: Understanding a limit from a table

Consider:

\[f(x)=x^2\]

What happens as \(x\) approaches 2?

x f ( x ) = x 2
1.9 3.61
1.99 3.9601
1.999 3.996001
2.001 4.004001
2.01 4.0401
2.1 4.41

As \(x\) becomes closer to 2 from either side, \(f(x)\) approaches 4.

Therefore:

\[\lim_{x\to2}x^2=4\]

For this particular function, the limit equals the actual function value \(f(2)=4\). However, this equality does not always hold for every function. A limit concerns the values near the target input, not necessarily the value at the target itself.

Approaching from the left and right

Suppose \(x\) approaches 2. Values approaching from the left include:

\[1.9,\quad1.99,\quad1.999\]

Values approaching from the right include:

\[2.1,\quad2.01,\quad2.001\]

For a two-sided limit to exist as a finite real number, the left-hand and right-hand limits must agree. This concept becomes important when determining whether a derivative exists at a point.

02.11.5 The Difference Quotient

The difference quotient measures the average rate of change over an interval expressed in terms of a small change \(h\). Suppose we start at an input \(x\). After changing the input by \(h\), the new input is:

\[x+h\]

The original output is:

\[f(x)\]

The new output is:

\[f(x+h)\]

Therefore, the change in output is:

\[f(x+h)-f(x)\]

The change in input is:

\[h\]

The difference quotient is:

\[\frac{f(x+h)-f(x)}{h},\qquad h\neq0\]
  • The value of \(h\) may be positive or negative.
  • A small positive \(h\) corresponds to moving slightly to the right of \(x\), while a small negative \(h\) corresponds to moving slightly to the left.
  • In this illustration, point \(P\) stays fixed while point \(Q\) moves closer to \(P\). As the positive value of \(h\) decreases, the secant line approaches the tangent line at \(P\).
  • This illustrates a right-hand approach. The ordinary derivative at an interior point requires the same finite limiting slope when approaching from both the left and the right.

Example 5: Finding the difference quotient of \(f(x)=x^2\)

Given:

\[f(x)=x^2\]

Find:

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

Step 1. Calculate \(f(x+h)\).

Substitute \(x+h\) into the function:

\[f(x+h)=(x+h)^2\]

Expand:

\[f(x+h)=x^2+2xh+h^2\]

Step 2. Subtract \(f(x)\).

\[f(x+h)-f(x) =x^2+2xh+h^2-x^2\]

Simplify:

\[f(x+h)-f(x)=2xh+h^2\]

Step 3. Divide by \(h\).

\[\frac{2xh+h^2}{h}\]

Factor:

\[\frac{h(2x+h)}{h}\]

Since \(h\neq0\), cancel \(h\):

\[2x+h\]

Final answer:

\[\frac{f(x+h)-f(x)}{h}=2x+h\]

This result is not yet the derivative. It is the average rate of change from \(x\) to \(x+h\).

Difference Quotient

Difference quotient =

f ( x + h ) − f ( x )

h

The input change h must be nonzero.

02.11.6 From the Difference Quotient to the Derivative

  1. We can now define the derivative mathematically.
  2. The difference quotient gives the slope of a secant line.
  3. If we let \(h\) approach zero, the two points move closer together.
  4. When the difference quotient approaches the same finite value from both sides, that value is the derivative.

Definition of the derivative

The derivative of \(f\) at \(x=a\) is:

\[f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]

provided the limit exists.

Here:

  • \(a\) is the input at which the derivative is evaluated.
  • \(h\) is a small nonzero change in the input.
  • \(f(a+h)-f(a)\) is the change in output.
  • The fraction gives the average rate of change.
  • The limit gives the instantaneous rate of change.

Example 6: Find the derivative of \(f(x)=x^2\) from first principles

We want to find the derivative at an arbitrary point \(x=a\).

Start with:

\[f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]

For:

\[f(x)=x^2\]

we have:

\[f(a+h)=(a+h)^2\]

and:

\[f(a)=a^2\]

Substitute:

\[f'(a)=\lim_{h\to0}\frac{(a+h)^2-a^2}{h}\]

Expand:

\[f'(a)=\lim_{h\to0}\frac{a^2+2ah+h^2-a^2}{h}\]

Simplify:

\[f'(a)=\lim_{h\to0}\frac{2ah+h^2}{h}\]

Factor \(h\):

\[f'(a)=\lim_{h\to0}(2a+h)\]

We canceled \(h\) only while \(h\neq0\). We are now calculating what the simplified expression approaches as \(h\) becomes arbitrarily close to zero.

Therefore:

\[f'(a)=2a\]

Because \(a\) can be any real input:

\[f'(x)=2x\]

Final answer: The derivative of \(x^2\) is \(2x\).

Example 7: Find the derivative at \(x=2\)

Using:

\[f'(x)=2x\]

Substitute \(x=2\):

\[f'(2)=2(2)=4\]

Therefore, the instantaneous rate of change at \(x=2\) is 4.

Exploring the limit numerically

For \(f(x)=x^2\) at \(x=2\):

\[\frac{f(2+h)-f(2)}{h}=4+h\]

Difference Quotient as h Approaches Zero

For f(x) = x² at x = 2, the difference quotient is 4 + h.

h 4 + h
1 5
0.5 4.5
0.1 4.1
0.01 4.01
−0.01 3.99
−0.1 3.9
−0.5 3.5

As h approaches 0 from either direction, 4 + h approaches 4. The derivative at x = 2 is therefore 4.

f ′( a ) =

lim h→0

f ( a + h ) − f ( a )

h

02.11.7 Derivative Notation

There are several common ways to represent derivatives. The most important notation for this course is:

\[f'(x)\]

This is read as f prime of x.

If:

\[f(x)=x^3\]

then:

\[f'(x)=3x^2\]

The derivative is itself a function that describes the instantaneous rate of change at different inputs.

Common Derivative Notation

Notation Meaning
f ′( x ) Derivative function
f ′( a ) Derivative evaluated at x = a
dy dx Derivative of y with respect to x
d dx f ( x ) Differentiate f ( x ) with respect to x
dy dx | x = a Derivative evaluated at x = a

Example 8: Understanding the notation

Suppose:

\[f(x)=x^3\]

Then:

\[f'(x)=3x^2\]

At \(x=2\):

\[f'(2)=3(2)^2\]
\[=12\]

Here, \(f'(x)\) is a formula, while \(f'(2)\) is a number. The original function value is:

\[f(2)=2^3=8\]

The derivative value is:

\[f'(2)=12\]

These are different quantities. Remember: \(f'(2)\) does not mean \(f(2)\), \(1/f(2)\), or \(f(x+1)\).

02.11.8 The Derivative of a Constant

A constant function has the form:

\[f(x)=c\]

where \(c\) is a fixed real number. For example:

\[f(x)=7\]

The function always produces the same output, regardless of its input. Its graph is a horizontal line. Because the output never changes, its instantaneous rate of change is zero.

Therefore:

\[\frac{d}{dx}(c)=0\]

Example 9: Differentiate a constant

Given:

\[f(x)=9\]

Find \(f'(x)\). Because 9 is constant:

\[f'(x)=0\]

Example 10: Derivative of a negative constant

Given:

\[g(x)=-15\]

The output remains −15 for every input. Therefore:

\[g'(x)=0\]

Why is the derivative zero?

Using the definition:

\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]

For a constant function, both outputs are equal to \(c\). Therefore:

\[f'(x)=\lim_{h\to0}\frac{c-c}{h}\]
\[=\lim_{h\to0}0=0\]

This confirms the constant rule.

02.11.9 The Power Rule

The power rule is one of the most important differentiation rules. It allows us to calculate derivatives of power functions without repeating the full limit calculation each time. For:

\[f(x)=x^n\]

the power rule states:

\[\frac{d}{dx}(x^n)=nx^{n-1}\]

at points where the power function is differentiable. For positive integer exponents, the rule applies at every real input. For negative and fractional exponents, domain restrictions must be considered.

How to apply the power rule

There are two main steps:

  1. Multiply by the original exponent.
  2. Reduce the exponent by 1.

For example:

\[\frac{d}{dx}(x^5)=5x^{5-1}\]

Therefore:

\[\frac{d}{dx}(x^5)=5x^4\]

Example 11: Differentiate \(x^4\)

Given:

\[f(x)=x^4\]

Apply the power rule:

\[f'(x)=4x^{4-1}\]

Therefore:

\[f'(x)=4x^3\]

Example 12: Differentiate \(x^7\)

Given:

\[f(x)=x^7\]

Then:

\[f'(x)=7x^6\]

Example 13: Differentiate \(x\)

We can write:

\[f(x)=x^1\]

Apply the power rule:

\[f'(x)=1x^{1-1}\]
\[=1x^0\]

Since \(x^0=1\) for \(x\neq0\), and the derivative of the identity function at zero is also 1:

\[f'(x)=1\]

for every real \(x\). This makes sense because the graph of \(y=x\) has constant slope 1.

Example 14: Differentiate a negative integer power

Given:

\[f(x)=x^{-2}\]

The domain excludes zero. Apply the power rule:

\[f'(x)=-2x^{-3}\]

Rewrite using positive exponents:

\[f'(x)=-\frac2{x^3}\]

Final answer:

\[-\frac2{x^3},\qquad x\neq0\]

Example 15: Differentiate a fractional power

Given:

\[f(x)=x^{1/2}=\sqrt{x}\]

Apply the power rule for \(x>0\):

\[f'(x)=\frac12x^{-1/2}\]

Rewrite:

\[f'(x)=\frac1{2\sqrt{x}}\]

Final answer:

\[f'(x)=\frac1{2\sqrt{x}},\qquad x>0\]

Although \(\sqrt{x}\) is defined at \(x=0\), it does not have a finite ordinary derivative there.

Example 16: Differentiate a function with a fractional exponent

Given:

\[f(x)=x^{3/2}\]

For \[
x>0
\] 

\[f'(x)=\frac32x^{1/2}\]

Therefore:

\[f'(x)=\frac32\sqrt{x}\]

At the domain endpoint \(x=0\), the right-hand difference quotient approaches 0. The ordinary two-sided derivative requires an open neighborhood and is not defined there under the standard convention used in this lesson.

The Power Rule

General Formula

d/dx ( x n )

= n x n −1

Multiply by the exponent, then subtract 1 from the exponent.

Original function Derivative
x 1
x 2 2 x
x 3 3 x 2
x 4 4 x 3
x 5 5 x 4
x −1 − x −2
x −2 −2 x −3
x 1/2 (1/2) x −1/2 , x > 0

For negative and fractional powers, respect the original function’s domain

and the points where its derivative exists.

02.11.10 The Constant-Multiple Rule

In many functions, a power of \(x\) is multiplied by a constant.

For example:

\[f(x)=5x^3\]

To differentiate this expression, we apply the constant-multiple rule.

If \(c\) is constant:

\[\frac{d}{dx}[cf(x)]=cf'(x)\]

This means the constant coefficient remains outside the differentiation operation.

Example 17: Differentiate \(5x^3\)

Given:

\[f(x)=5x^3\]

First differentiate \(x^3\):

\[\frac{d}{dx}(x^3)=3x^2\]

Then multiply by 5:

\[f'(x)=5(3x^2)\]

Therefore:

\[f'(x)=15x^2\]

Example 18: Negative coefficient

Given:

\[f(x)=-4x^5\]

Apply the constant-multiple and power rules:

\[f'(x)=-4(5x^4)\]

Therefore:

\[f'(x)=-20x^4\]

Example 19: Fractional coefficient

Given:

\[f(x)=\frac32x^4\]

Differentiate:

\[f'(x)=\frac32(4x^3)\]
\[f'(x)=6x^3\]

Important: The coefficient multiplies the new exponent. It does not disappear.

02.11.11 The Sum and Difference Rules

Many functions contain several terms added or subtracted together.

For example:

\[f(x)=x^3+2x^2-5x+7\]

We can differentiate each term separately.

Sum rule

If \(f\) and \(g\) are differentiable:

\[\frac{d}{dx}[f(x)+g(x)] =f'(x)+g'(x)\]

Difference rule

Similarly:

\[\frac{d}{dx}[f(x)-g(x)] =f'(x)-g'(x)\]

Example 20: Differentiating a sum

Given:

\[f(x)=x^3+x^2\]

Differentiate each term:

\[f'(x)=3x^2+2x\]

Final answer:

\[3x^2+2x\]

Example 21: Differentiating a difference

Given:

\[f(x)=x^4-3x^2\]

Differentiate term by term:

\[f'(x)=4x^3-6x\]

Example 22: A function with several terms

Given:

\[f(x)=3x^4-5x^3+2x-7\]

Differentiate each term separately.

First term:

\[\frac{d}{dx}(3x^4)=12x^3\]

Second term:

\[\frac{d}{dx}(-5x^3)=-15x^2\]

Third term:

\[\frac{d}{dx}(2x)=2\]

Fourth term:

\[\frac{d}{dx}(-7)=0\]

Combine:

\[f'(x)=12x^3-15x^2+2\]

Final answer:

\[12x^3-15x^2+2\]

Fundamental Differentiation Rules

Rule Formula
Constant (c)’ = 0
Identity (x)’ = 1
Power (x n )’ = nx n-1
Constant multiple (cf)’ = cf’
Sum (f + g)’ = f’ + g’
Difference (f – g)’ = f’ – g’

These formulas assume the relevant functions are differentiable.

02.11.12 Differentiating Polynomial Functions

A polynomial function is a function composed of a finite sum of nonnegative integer powers of \(x\), multiplied by constant coefficients.

For example:

\[f(x)=4x^3-2x^2+7x-5\]

Every polynomial is differentiable at every real input. We can therefore use the power, constant-multiple, sum, and difference rules to find its derivative.

General form

A polynomial has the form:

\[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\]

Its derivative is:

\[f'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1\]

The constant term \(a_0\) disappears because its derivative is zero.

Example 23: Differentiate a quadratic polynomial

Given:

\[f(x)=5x^2-4x+6\]

Differentiate each term.

\[\frac{d}{dx}(5x^2)=10x\]
\[\frac{d}{dx}(-4x)=-4\]
\[\frac{d}{dx}(6)=0\]

Therefore:

\[f'(x)=10x-4\]

Example 24: Differentiate a cubic polynomial

Given:

\[g(x)=2x^3-3x^2+8x-10\]

Differentiate:

\[g'(x)=6x^2-6x+8\]

Notice that the constant term −10 disappears.

Example 25: Differentiate a fourth-degree polynomial

Given:

\[h(x)=-2x^4+5x^3-7x+9\]

Differentiate term by term:

\[h'(x)=-8x^3+15x^2-7\]

Final answer:

\[-8x^3+15x^2-7\]

Example 26: Find a derivative at a particular input

Given:

\[f(x)=2x^3-5x^2+4x+1\]

Find \(f'(2)\).

Step 1. Differentiate.

\[f'(x)=6x^2-10x+4\]

Step 2. Substitute \(x=2\).

\[f'(2)=6(2)^2-10(2)+4\]
\[=24-20+4\]
\[=8\]

Final answer:

\[f'(2)=8\]

This means the instantaneous rate of change at \(x=2\) is 8.

Example 27: Find the input where a derivative equals zero

Consider:

\[f(x)=x^2-6x+5\]

Find the value of \(x\) for which:

\[f'(x)=0\]

Step 1. Differentiate.

\[f'(x)=2x-6\]

Step 2. Set the derivative equal to zero.

\[2x-6=0\]
\[2x=6\]
\[x=3\]

Final answer:

\[x=3\]

At this point, the tangent is horizontal.

A zero derivative identifies a stationary point, but does not by itself establish whether the point is a maximum or minimum. This distinction is explored later in Lesson 02.13.

02.11.13 A Simple Application: Instantaneous Velocity

Derivatives are widely used in physics to describe motion.

If an object’s position is given by:

\[s(t)\]

where \(t\) represents time, its instantaneous velocity is:

\[v(t)=s'(t)\]

This means velocity is the derivative of position with respect to time.

Example 28: Velocity from a position function

Suppose a moving object’s position is:

\[s(t)=t^2+3t\]

where position is measured in meters and time in seconds. Find the instantaneous velocity at \(t=2\) seconds.

Step 1. Differentiate the position function.

\[v(t)=s'(t)=2t+3\]

Step 2. Substitute \(t=2\).

\[v(2)=2(2)+3\]
\[=7\]

Final answer:

\[v(2)=7\text{ m/s}\]

Compare with average velocity

Over the interval from \(t=1\) to \(t=2\):

\[s(1)=1^2+3(1)=4\]
\[s(2)=2^2+3(2)=10\]

The average velocity is:

\[v_{\text{avg}}=\frac{10-4}{2-1}=6\text{ m/s}\]

Compare:

Quantity Value
Average velocity from 1 to 2 seconds 6 m/s
Instantaneous velocity at 2 seconds 7 m/s
  • This demonstrates the distinction between change over an interval and change at a particular moment.
  • Important: Velocity includes direction and may be negative. Speed is the magnitude of velocity in one-dimensional motion.
  • Also note that the units of a derivative are the output units divided by the input units. For position measured in meters and time measured in seconds, velocity is measured in meters per second.

02.11.14 When Does a Derivative Not Exist?

  • Not every function is differentiable at every point of its domain.
  • A derivative may fail to exist when the difference quotient does not approach a single finite value.
  • One important example is a graph with a sharp corner.

Example 29: The absolute-value function

Consider:

\[f(x)=|x|\]

We will examine the derivative at:

\[x=0\]

The derivative definition gives:

\[f'(0)=\lim_{h\to0}\frac{|h|-0}{h}\]

Therefore:

\[f'(0)=\lim_{h\to0}\frac{|h|}{h}\]

When

\[
h>0
\]

\[|h|=h\]

so:

\[\frac{|h|}{h}=1\]

When

\[
h<0
\]

\[|h|=-h\]

so:

\[\frac{|h|}{h}=-1\]

The right-hand limit is 1, while the left-hand limit is −1. These do not agree.

Therefore:

\[f'(0)\text{ does not exist}\]

The graph of \(y=|x|\) has a sharp corner at the origin.

What can prevent differentiability?

Common introductory examples include:

Situations Where a Derivative May Not Exist

Situation Example Explanation
Sharp corner |x| at x = 0 Left and right slopes differ
Discontinuity 1/x at x = 0 Function is undefined
Vertical tangent Cube root of x at x = 0 No finite derivative

02.11.15 Common Mistakes to Avoid

Mistake 1: Confusing a function with its derivative

For:

\[f(x)=x^2\]

the function value at 3 is:

\[f(3)=9\]

But the derivative value is:

\[f'(3)=6\]

These represent different quantities.

Mistake 2: Treating an average rate as an instantaneous rate

The average slope over an interval does not generally equal the slope at its endpoints. Use the derivative when the problem asks about the instantaneous rate at a specific input.

Mistake 3: Substituting \(h=0\) too early

Consider:

\[\frac{(x+h)^2-x^2}{h}\]

Directly substituting \(h=0\) produces:

\[\frac00\]

which is undefined. Instead, simplify the expression for \(h\neq0\), then take the limit.

Mistake 4: Forgetting to subtract 1 from the exponent

Incorrect:

\[\frac{d}{dx}(x^5)=5x^5\]

Correct:

\[\frac{d}{dx}(x^5)=5x^4\]

Mistake 5: Forgetting to multiply by the original exponent

Incorrect:

\[\frac{d}{dx}(4x^3)=4x^2\]

Correct:

\[\frac{d}{dx}(4x^3)=12x^2\]

Mistake 6: Differentiating constants incorrectly

For:

\[f(x)=x^2+7\]

the derivative is:

\[f'(x)=2x\]

not \(2x+7\).

Mistake 7: Treating differentiation as ordinary subtraction

The expression:

\[f'(x)\]

does not mean \(f(x)-1\). The prime symbol represents differentiation.

Mistake 8: Applying a power rule without checking the domain

For:

\[f(x)=\sqrt{x}\]

the formula:

\[f'(x)=\frac1{2\sqrt{x}}\]

applies for \[x>0\]

It does not apply at \(x=0\).

Mistake 9: Assuming a zero derivative guarantees a maximum or minimum

A zero derivative identifies a stationary point.

For example:

\[f(x)=x^3\]

has:

\[f'(x)=3x^2\]

and:

\[f'(0)=0\]

but the function continues increasing through zero, so the point is not a local maximum or minimum.

Mistake 10: Incorrectly differentiating a product

The sum rule does not mean:

\[(fg)’=f’g’\]

That identity is generally false.

For example, if \(f(x)=x\) and \(g(x)=x\), then \(fg=x^2\), whose derivative is \(2x\), not \(1\). The product rule is a separate topic and is not needed for the basic polynomial problems in this lesson.

02.11.16 General Strategy for Basic Derivatives

Step 1: Identify the function

Constant, power, polynomial, or another function?

↓

Step 2: Check what the question asks

Derivative formula, derivative value, or average rate?

↓

Step 3: Choose the correct rule

Definition, constant, power, sum, or difference.

↓

Step 4: Differentiate

Apply the rule carefully to each term.

↓

Step 5: Simplify

Combine like terms and rewrite if needed.

↓

Step 6: Evaluate if requested

Substitute the given input.

↓

Step 7: Verify

Check calculation, domain, and units.

02.11.17 Lesson Summary

Review the difference between a term’s value and its instantaneous rate of change; secant and tangent lines; difference quotients and limits; and the constant, power, constant-multiple, sum and difference rules.

Essential formulas

\[\text{Average rate}=\frac{f(b)-f(a)}{b-a}\]
\[
\frac{f(x+h)-f(x)}{h},\qquad h\neq 0
\]
\[\frac{d}{dx}(x^n)=nx^{n-1}\]
\[
f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}
\]

Lesson completion standard: Explain how the difference quotient leads to a derivative; differentiate elementary polynomials; evaluate a derivative at a specified input; and identify basic cases of non-differentiability.

Essential Formula Reference

Concept Formula
Average rate of change f ( b ) − f ( a ) b − a
Difference quotient f ( x + h ) − f ( x ) h
Derivative at a f ′( a ) = lim h → 0 f ( a + h ) − f ( a ) h
Derivative of a constant ( c )′ = 0
Power rule ( x n )′ = n x n − 1
Constant multiple ( cf )′ = cf ′
Sum rule ( f + g )′ = f ′ + g ′
Difference rule ( f − g )′ = f ′ − g ′
Instantaneous velocity v ( t ) = s ′( t )

Lesson completion standard, students should be able to explain how the

  • difference quotient leads to the derivative,
  • differentiate elementary polynomials, and
  • evaluate the resulting derivative at a specified input.