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MathematicsHigh schoolEasy

Derivative from the definition

Using the definition of the derivative, find f′(x)f'(x) for the function

f(x)=3x2−x.f(x) = 3x^2 - x.
Hintshow

Write out the difference quotient f(x+h)−f(x)h\dfrac{f(x + h) - f(x)}{h}, expand (x+h)2(x + h)^2 and simplify the expression before taking the limit.

Solutiontry it on your own first

Step 1. Write down the definition of the derivative

We use the formula:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

So we need two expressions: f(x+h)f(x + h) and f(x)f(x).

Step 2. Work out f(x + h)

Our function is:

f(x)=3x2−xf(x) = 3x^2 - x

Replace xx with x+hx + h:

f(x+h)=3(x+h)2−(x+h)f(x + h) = 3(x + h)^2 - (x + h)

Expand the square:

(x+h)2=x2+2xh+h2(x + h)^2 = x^2 + 2xh + h^2

So:

f(x+h)=3(x2+2xh+h2)−x−h\begin{aligned} f(x + h) &= 3(x^2 + 2xh + h^2) \\ &\quad - x - h \end{aligned}

Multiply the bracket by 3:

f(x+h)=3x2+6xh+3h2−x−h\boxed{ \begin{aligned} f(x + h) &= 3x^2 + 6xh + 3h^2 \\ &\quad - x - h \end{aligned} }

Step 3. Work out the difference f(x + h) − f(x)

We subtract the whole of f(x)=3x2−xf(x) = 3x^2 - x, so we put it in brackets:

f(x+h)−f(x)=3x2+6xh+3h2−x−h−(3x2−x)\begin{aligned} & f(x + h) - f(x) \\ & = 3x^2 + 6xh + 3h^2 \\ & \quad - x - h - (3x^2 - x) \end{aligned}

Remove the brackets - the minus sign in front changes the sign of both terms:

f(x+h)−f(x)=3x2+6xh+3h2−x−h−3x2+x\begin{aligned} & f(x + h) - f(x) \\ & = 3x^2 + 6xh + 3h^2 \\ & \quad - x - h - 3x^2 + x \end{aligned}

Collect like terms: 3x2−3x2=03x^2 - 3x^2 = 0 and −x+x=0-x + x = 0:

f(x+h)−f(x)=6xh+3h2−h\boxed{ \begin{aligned} & f(x + h) - f(x) \\ & = 6xh + 3h^2 - h \end{aligned} }

Step 4. Divide the difference by h

Put the difference into the difference quotient:

f(x+h)−f(x)h=6xh+3h2−hh\frac{f(x + h) - f(x)}{h} = \frac{6xh + 3h^2 - h}{h}

Take hh out as a common factor in the numerator:

f(x+h)−f(x)h=h(6x+3h−1)h\frac{f(x + h) - f(x)}{h} = \frac{h(6x + 3h - 1)}{h}

For h≠0h \neq 0, cancel the hh:

f(x+h)−f(x)h=6x+3h−1\frac{f(x + h) - f(x)}{h} = 6x + 3h - 1

Step 5. Take the limit as h → 0

We have:

f′(x)=lim⁡h→0(6x+3h−1)f'(x) = \lim_{h \to 0} (6x + 3h - 1)

There is no hh in the denominator any more, so we can put h=0h = 0:

f′(x)=6x+3⋅0−1=6x−1f'(x) = 6x + 3 \cdot 0 - 1 = 6x - 1

Answer

f′(x)=6x−1\boxed{f'(x) = 6x - 1}

The derivative of f(x)=3x2−xf(x) = 3x^2 - x is f′(x)=6x−1f'(x) = 6x - 1. It matches the differentiation rules: (3x2)′=3⋅2x=6x(3x^2)' = 3 \cdot 2x = 6x and (−x)′=−1(-x)' = -1.

The theory behind this exerciseCourse: Derivatives from scratch

Sources

We checked the facts on this page against these sources. All of them may be reused freely: they carry an open licence or are not covered by copyright.

  1. Active Calculus (Boelkins) - 1.3 The derivative of a function at a point (limit of the difference quotient) · CC BY-SA 4.0
  2. Active Calculus (Boelkins) - 2.1 Elementary derivative rules (power rule) · CC BY-SA 4.0
  3. Dale Hoffman - Contemporary Calculus, 2.1 The Definition of Derivative (PDF) · CC BY 3.0 US
  4. Calaway, Hoffman and Lippman - Applied Calculus, 2.2 The Derivative and 2.3 Power and Sum Rules for Derivatives (PDF) · CC BY 3.0 US
  5. Wikibooks - Calculus/Differentiation/Differentiation Defined (difference quotient, definition of the derivative) · CC BY-SA 4.0
  6. Wikiversity - Mathematics for Applied Sciences (Osnabrück 2023-2024), Part I, Lecture 14: Differentiability (difference quotient, the derivative of x² from the definition) · CC BY-SA 4.0

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