# Derivative Rules

See how to find the derivative of f(x)

by using the derivative rules.

20 examples and their solutions.

## Derivarive: Definition

### Definition

f'(a) = limx → af(x) - f(a)x - a

### Example

f(x) = x

f'(2) = ?

(Use the definition of the derivative.)

Solution ^{2}- 3x + 1f'(2) = ?

(Use the definition of the derivative.)

f'(2) = limh → 0f(2 + h) - f(2)h

= limh → 0[(2 + h)

= limh → 02

= limh → 04 + 4h + h

= limh → 0h

= limh → 0h + 11

= 0 + 11 - [2]

= 11

= 1

= limh → 0[(2 + h)

^{2}- 3(2 + h) + 1] - [2^{2}- 3⋅2 + 1]h= limh → 02

^{2}+ 2⋅2⋅h + h^{2}- 6 - 3h + 1 - [4 - 6 + 1]h - [1]= limh → 04 + 4h + h

^{2}- 6 - 3h + 1 - 4 + 6 -1h= limh → 0h

^{2}+ hh= limh → 0h + 11

= 0 + 11 - [2]

= 11

= 1

Graph

f'(2) = 1: the slope of y = f(x) is 1 at x = 2.

[1]

[2]

Close

## Differentiable

### Definition

limx → a

^{-}f'(x) = limx → a

^{+}f'(x)

Differentiable at x = a: y = f'(x) exist at x = a.

→ (left-hand slope) = (right-hand slope)

→ y = f(x) is a smooth curve at x = a.

How to find out:

1. Show that f(x) is continuous at x = a.

(left-hand limit) = (right-hand limit) = f(a)

2. (left-hand derivative) = (right-hand derivative)

### Example

f(x) = {x

-x

Determine whether f(x) is differentiable at x = 1.

Solution ^{2}+ 2 (x < 1)-x

^{2}+ 4x (x ≥ 1)Determine whether f(x) is differentiable at x = 1.

limx → 1

= limx → 1

= (1

= 1

= 1 + 2

= 3

limx → 1

= limx → 1

= -(1

= -1

= -1 + 4

= 3

f(1) = -1

= -1 + 4

= 3

limx → 1

∴ f(x) is continuous at x = 1.

limx → 1

= limx → 1

= limx → 1

= 2⋅(1

= 2⋅1

= 2

limx → 1

= limx → 1

= limx → 1

= -2⋅(1

= -2⋅1 + 4

= -2 + 4

= 2

limx → 1

∴ f(x) is differentiable at x = 1.

^{-}f(x) - [1]= limx → 1

^{-}(x^{2}+ 2)= (1

^{-})^{2}+ 2= 1

^{2}+ 2= 1 + 2

= 3

limx → 1

^{+}f(x)= limx → 1

^{+}(-x^{2}+ 4x)= -(1

^{+})^{2}+ 4⋅(1^{+})= -1

^{2}+ 4⋅1= -1 + 4

= 3

f(1) = -1

^{2}+ 4⋅1= -1 + 4

= 3

limx → 1

^{-}f(x) = limx → 1^{+}f(x) = f(1)∴ f(x) is continuous at x = 1.

limx → 1

^{-}f'(x) - [2]= limx → 1

^{-}(2x^{1}+ 0)= limx → 1

^{-}2x= 2⋅(1

^{-})= 2⋅1

= 2

limx → 1

^{+}f'(x)= limx → 1

^{+}(-2x^{1}+ 4)= limx → 1

^{+}(-2x + 4)= -2⋅(1

^{+}) + 4= -2⋅1 + 4

= -2 + 4

= 2

limx → 1

^{-}f'(x) = limx → 1^{+}f'(x)∴ f(x) is differentiable at x = 1.

Graph

y = f(x) is differentiable at x = 1.

→ y = f(x) is a smooth curve at x = 1.

→ y = f(x) is a smooth curve at x = 1.

[1]

First show that f(x) is continuous at x = 1.

(left-hand limit) = (right-hand limit) = f(1)

(left-hand limit) = (right-hand limit) = f(1)

[2]

Show that (left-hand derivative) = (right-hand derivative).

Close

### Example

f(x) = |x|

Determine whether f(x) is differentiable at x = 0.

Solution Determine whether f(x) is differentiable at x = 0.

f(x) = {-x (x < 0)

x (x ≥ 0) - [1]

limx → 0

= limx → 0

= -(0

= -0

= 0

limx → 0

= limx → 0

= 0

= 0

f(0) = 0

limx → 0

∴ f(x) is continuous at x = 0.

limx → 0

= limx → 0

= -1

limx → 0

= limx → 0

= 1

limx → 0

∴ f(x) is not differentiable at x = 0.

x (x ≥ 0) - [1]

limx → 0

^{-}f(x)= limx → 0

^{-}(-x)= -(0

^{-})= -0

= 0

limx → 0

^{+}f(x)= limx → 0

^{+}x= 0

^{+}= 0

f(0) = 0

limx → 0

^{-}f(x) = limx → 0^{+}f(x) = f(0)∴ f(x) is continuous at x = 0.

limx → 0

^{-}f'(x)= limx → 0

^{-}(-1)= -1

limx → 0

^{+}f'(x)= limx → 0

^{+}1= 1

limx → 0

^{-}f'(x) ≠ limx → 0^{+}f'(x)∴ f(x) is not differentiable at x = 0.

Graph

y = f(x) is not differentiable at x = 0.

→ y = f(x) is not a smooth curve at x = 0.

→ y = f(x) is not a smooth curve at x = 0.

Close

## Derivative Function

### Definition

f'(x) = limh → 0f(x + h) - f(x)x

A derivative function y = f'(x) showsthe slope of y = f(x) for each x.

### How to Write

f'(x), y', dydx, ddxf(x)

These are the ways to write the derivative function. ## Derivative of a Constant

### Formula

[C]' = 0

The graph of y = C is a horizontal line.→ (slope) = 0

→ y' = 0

→ [C]' = 0

Linear Equation (Two Variables)

## Derivative of mx

### Formula

## Derivative of a⋅f(x) + b⋅g(x)

### Formula

y = af(x) + bg(x)

→ y' = af'(x) + bg'(x)

→ y' = af'(x) + bg'(x)

## Derivarive of x^{n}

### Formula

[x

n: real number ^{n}]' = nx^{n - 1}### Example

y = x

y' = ?

Solution ^{3}y' = ?

y = x

y' = 3x

^{3}y' = 3x

^{2}Close

### Example

f(x) = 2x

f'(x) = ?

Solution ^{7}- 5x + 3f'(x) = ?

f(x) = 2x

f'(x) = 2⋅7x

= 14x

^{7}- 5x + 3f'(x) = 2⋅7x

^{6}- 5 + 0= 14x

^{6}- 5Close

### Example

y = 6x

y' = ?

Solution ^{6}- 3x^{3}+ 2x^{2}- 1x^{3}y' = ?

y = 6x

= 6x

y' = 6⋅3x

= 18x

^{6}- 3x^{3}+ 2x^{2}- 1x^{3}= 6x

^{3}- 3 + 2x^{-1}- x^{-3}- [1]y' = 6⋅3x

^{2}- 0 + 2⋅-1x^{-2}- (-3)x^{-4}= 18x

^{2}- 2x^{2}+ 3x^{4}Close

### Example

y = √x

y' = ?

Solution y' = ?

### Example

y = 1

y' = ?

Solution ^{4}√x^{3}y' = ?

y = 1

= 1x

= x

y' = -34x

= -34x

= -34⋅x⋅x

= -34x⋅

^{4}√x^{3}= 1x

^{34}= x

^{-34}y' = -34x

^{-34 - 1}= -34x

^{34 + 1}= -34⋅x⋅x

^{34}= -34x⋅

^{4}√x^{3}Close

### Example

## Derivative of f(x)g(x)

### Formula

[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)

### Example

y = (2x

y' = ?

Solution ^{3}- 5)(4x^{2}+ x)y' = ?

y = (2x

y' = (2⋅3x

= 6x

= 24x

= 40x

^{3}- 5)(4x^{2}+ x)y' = (2⋅3x

^{2}- 0)(4x^{2}+ x) + (2x^{3}- 5)(4⋅2x^{1}+ 1)= 6x

^{2}(4x^{2}+ x) + (2x^{3}- 5)(8x + 1)= 24x

^{4}+ 6x^{3}+ 16x^{4}+ 2x^{3}- 40x - 5 - [1]= 40x

^{4}+ 8x^{3}- 40x - 5[1]

Close

### Example

f(x) = (x

f(1) = ?

Solution ^{5}- 2x)(7x^{2}+ 3)f(1) = ?

y = (x

y' = (5x

= (5x

f'(1) = (5⋅1

= (5⋅1 - 2)(7⋅1 + 3) + (1 - 2)⋅14

= (5 - 2)(7 + 3) - 1⋅14

= 3⋅10 - 14

= 30 - 14

= 16

^{5}- 2x)(7x^{2}+ 3)y' = (5x

^{4}- 2)(7x^{2}+ 3) + (x^{5}- 2x)(7⋅2x^{1}+ 0)= (5x

^{4}- 2)(7x^{2}+ 3) + (x^{5}- 2x)⋅14xf'(1) = (5⋅1

^{4}- 2)(7⋅1^{2}+ 3) + (1^{5}- 2⋅1)⋅14⋅1= (5⋅1 - 2)(7⋅1 + 3) + (1 - 2)⋅14

= (5 - 2)(7 + 3) - 1⋅14

= 3⋅10 - 14

= 30 - 14

= 16

Close

## Derivative of 1g(x)

### Formula

(1g(x))' = -g'(x)(g(x))

^{2}### Example

y = 1x

y' = ?

Solution ^{3}+ 2xy' = ?

y = 1x

y' = -3x

^{3}+ 2xy' = -3x

^{2}+ 2(x^{3}+ 2x)^{2}Close

y = 1x

= (x

y' = -1⋅(x

= -3x

^{3}+ 2x= (x

^{3}+ 2x)^{-1}- [1]y' = -1⋅(x

^{3}+ 2x)^{-2}⋅(3x^{2}+ 2) - [2]= -3x

^{2}+ 2(x^{3}+ 2x)^{2}Close

## Derivative of f(x)g(x)

### Formula

(f(x)g(x))' = -f'(x)g(x) - f(x)g'(x)(g(x))

^{2}### Example

y = 4xx

y' = ?

Solution ^{2}- 3y' = ?

y = 4xx

y' = -4⋅(x

= -4⋅(x

= -4x

= --4x

= 4x

^{2}- 3y' = -4⋅(x

^{2}- 3) - 4x⋅(2x^{1}- 0)(x^{2}- 3)^{2}= -4⋅(x

^{2}- 3) - 4x⋅2x(x^{2}- 3)^{2}= -4x

^{2}- 12 - 8x^{2}(x^{2}- 3)^{2}= --4x

^{2}- 12(x^{2}- 3)^{2}= 4x

^{2}+ 12(x^{2}- 3)^{2}Close

## Derivative of g(f(x))

### Formula

[g(f(x))]' = g'(f(x))⋅f'(x)

1. Think f(x) as a wholeand differentiate g(f(x)). → g'(f(x))

2. Differentiate f(x). → f'(x)

### Example

y = (2x

y' = ?

Solution ^{2}- 1)^{8}y' = ?

y = (2x

y' = 8⋅(2x

= 8(2x

= 32x(2x

^{2}- 1)^{8}y' = 8⋅(2x

^{2}- 1)^{7}⋅(2⋅2x^{1}- 0)= 8(2x

^{2}- 1)^{7}⋅4x= 32x(2x

^{2}- 1)^{7}Close

## Derivative of f(x, y) = 0

### Implicit Function

f(x, y) = 0

An implicit function is a functionthat cannot be changed to [y = ...] or [x = ...].

(= x and y are mixed.)

### Formula

f(x) → f'(x)

g(y) → g'(y)y'

When differenciating the y term g(y),g(y) → g'(y)y'

first differentiate g(y), g'(y),

then write y'.

y' = dy/dx

Derivative of g(f(x))

### Example

x

dydx = ?

Solution ^{2}+ y^{2}= 1dydx = ?

x

2x

2x + 2yy' = 0

x + yy' = 0

yy' = -x

y' = -xy

^{2}+ y^{2}= 12x

^{1}+ 2y^{1}y' = 02x + 2yy' = 0

x + yy' = 0

yy' = -x

y' = -xy

Close

### Example

x

dydx at (1, 1)?

Solution ^{3}+ xy^{2}- 2y^{3}+ 2 = 0dydx at (1, 1)?

x

3x

3x

2xyy' - 6y

y'(2xy - 6y

y' = -3x

[y']

= -3⋅1 - 12 - 6⋅1

= -3 - 12 - 6

= -4-4

= 1

^{3}+ xy^{2}- 2y^{3}+ 2 = 03x

^{2}+ 1⋅y^{2}+ x⋅2y^{1}y' - 2⋅3y^{2}y' + 0 = 0 - [1]3x

^{2}+ y^{2}+ 2xyy' - 6y^{2}y' = 02xyy' - 6y

^{2}y' = -3x^{2}- y^{2}y'(2xy - 6y

^{2}) = -3x^{2}- y^{2}y' = -3x

^{2}- y^{2}2xy - 6y^{2}[y']

_{(x, y) = (1, 1)}= -3⋅1^{2}- 1^{2}2⋅1⋅1 - 6⋅1^{2}= -3⋅1 - 12 - 6⋅1

= -3 - 12 - 6

= -4-4

= 1

[1]

Close

## Derivative of a Parametric Function

### Formula

dydx = dydt dxdt

### Example

x = t

y = t

dydx at t = 1?

Solution ^{3}- 2ty = t

^{2}+ 1dydx at t = 1?

x = t

y = t

dxdt = 3t

dydt = 2t

= 2t

dydx = dydt dxdt

= 2t3t

[dydx]

= 23⋅1 - 2

= 23 - 2

= 21

= 2

^{3}- 2ty = t

^{2}+ 1dxdt = 3t

^{2}- 2dydt = 2t

^{1}+ 0= 2t

dydx = dydt dxdt

= 2t3t

^{2}- 2[dydx]

_{t = 1}= 2⋅13⋅1^{2}- 2= 23⋅1 - 2

= 23 - 2

= 21

= 2

Close

## Derivative of an Inverse Function

### Definition

dxdy

When finding the inverse function,x and y are switched.

So the derivative of an inverse function is dx/dy.

### Formula

dxdy = 1 dydx

### Example

y = x

dxdy = ?

Solution ^{5}- x + 8dxdy = ?

y = x

dydx = 5x

= 5x

dxdy = 15x

^{5}- x + 8dydx = 5x

^{4}- 1 + 0= 5x

^{4}- 1dxdy = 15x

^{4}- 1Close

### Example

y = x

dxdy at y = 3?

Solution ^{3}+ 2dxdy at y = 3?

y = x

dydx = 3x

= 3x

dxdy = 13x

3 = x

1 = x

x

x = 1

dxdy = 13⋅1

= 13

^{3}+ 2dydx = 3x

^{2}+ 0= 3x

^{2}dxdy = 13x

^{2}3 = x

^{3}+ 2 - [1]1 = x

^{3}x

^{3}= 1x = 1

dxdy = 13⋅1

^{2}= 13

[1]

The goal is to find dx/dy at y = 3.

dx/dy = 1/3x

→ Find the x value when y = 3.

→ Put y = 3 into y = x

dx/dy = 1/3x

^{2}→ Find the x value when y = 3.

→ Put y = 3 into y = x

^{3}+ 2.Close

## Second Derivative

### How to Write

f''(x), y'', d

These are the ways to write the second derivative.^{2}ydx^{2}, d^{2}dx^{2}f(x)To find the second derivative,

differentiate f(x) twice.

Inflection Point

### Example

f(x) = x

f''(x) = ?

Solution ^{5}- 7x^{2}- 8x + 1f''(x) = ?

f(x) = x

f'(x) = 5x

= 5x

f''(x) = 5⋅4x

= 20x

^{5}- 7x^{2}- 8x + 1f'(x) = 5x

^{4}- 7⋅2x^{1}- 8 + 0= 5x

^{4}- 14x - 8f''(x) = 5⋅4x

^{3}- 14 - 0= 20x

^{3}- 14Close