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) = x2 - 3x + 1
f'(2) = ?
(Use the definition of the derivative.)
Solution f'(2) = ?
(Use the definition of the derivative.)
f'(2) = limh → 0f(2 + h) - f(2)h
= limh → 0[(2 + h)2 - 3(2 + h) + 1] - [22 - 3⋅2 + 1]h
= limh → 022 + 2⋅2⋅h + h2 - 6 - 3h + 1 - [4 - 6 + 1]h - [1]
= limh → 04 + 4h + h2 - 6 - 3h + 1 - 4 + 6 -1h
= limh → 0h2 + hh
= limh → 0h + 11
= 0 + 11 - [2]
= 11
= 1
= limh → 0[(2 + h)2 - 3(2 + h) + 1] - [22 - 3⋅2 + 1]h
= limh → 022 + 2⋅2⋅h + h2 - 6 - 3h + 1 - [4 - 6 + 1]h - [1]
= limh → 04 + 4h + h2 - 6 - 3h + 1 - 4 + 6 -1h
= limh → 0h2 + 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) = {x2 + 2 (x < 1)
-x2 + 4x (x ≥ 1)
Determine whether f(x) is differentiable at x = 1.
Solution -x2 + 4x (x ≥ 1)
Determine whether f(x) is differentiable at x = 1.
limx → 1-f(x) - [1]
= limx → 1-(x2 + 2)
= (1-)2 + 2
= 12 + 2
= 1 + 2
= 3
limx → 1+f(x)
= limx → 1+(-x2 + 4x)
= -(1+)2 + 4⋅(1+)
= -12 + 4⋅1
= -1 + 4
= 3
f(1) = -12 + 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-(2x1 + 0)
= limx → 1-2x
= 2⋅(1-)
= 2⋅1
= 2
limx → 1+f'(x)
= limx → 1+(-2x1 + 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.
= limx → 1-(x2 + 2)
= (1-)2 + 2
= 12 + 2
= 1 + 2
= 3
limx → 1+f(x)
= limx → 1+(-x2 + 4x)
= -(1+)2 + 4⋅(1+)
= -12 + 4⋅1
= -1 + 4
= 3
f(1) = -12 + 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-(2x1 + 0)
= limx → 1-2x
= 2⋅(1-)
= 2⋅1
= 2
limx → 1+f'(x)
= limx → 1+(-2x1 + 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-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.
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 xn
Formula
[xn]' = nxn - 1
n: real number Example
y = x3
y' = ?
Solution y' = ?
y = x3
y' = 3x2
y' = 3x2
Close
Example
f(x) = 2x7 - 5x + 3
f'(x) = ?
Solution f'(x) = ?
f(x) = 2x7 - 5x + 3
f'(x) = 2⋅7x6 - 5 + 0
= 14x6 - 5
f'(x) = 2⋅7x6 - 5 + 0
= 14x6 - 5
Close
Example
y = 6x6 - 3x3 + 2x2 - 1x3
y' = ?
Solution y' = ?
y = 6x6 - 3x3 + 2x2 - 1x3
= 6x3 - 3 + 2x-1 - x-3 - [1]
y' = 6⋅3x2 - 0 + 2⋅-1x-2 - (-3)x-4
= 18x2 - 2x2 + 3x4
= 6x3 - 3 + 2x-1 - x-3 - [1]
y' = 6⋅3x2 - 0 + 2⋅-1x-2 - (-3)x-4
= 18x2 - 2x2 + 3x4
Close
Example
Example
y = 14√x3
y' = ?
Solution y' = ?
y = 14√x3
= 1x34
= x-34
y' = -34x-34 - 1
= -34x34 + 1
= -34⋅x⋅x34
= -34x⋅4√x3
= 1x34
= x-34
y' = -34x-34 - 1
= -34x34 + 1
= -34⋅x⋅x34
= -34x⋅4√x3
Close
Example
Derivative of f(x)g(x)
Formula
[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)
Example
y = (2x3 - 5)(4x2 + x)
y' = ?
Solution y' = ?
y = (2x3 - 5)(4x2 + x)
y' = (2⋅3x2 - 0)(4x2 + x) + (2x3 - 5)(4⋅2x1 + 1)
= 6x2(4x2 + x) + (2x3 - 5)(8x + 1)
= 24x4 + 6x3 + 16x4 + 2x3 - 40x - 5 - [1]
= 40x4 + 8x3 - 40x - 5
y' = (2⋅3x2 - 0)(4x2 + x) + (2x3 - 5)(4⋅2x1 + 1)
= 6x2(4x2 + x) + (2x3 - 5)(8x + 1)
= 24x4 + 6x3 + 16x4 + 2x3 - 40x - 5 - [1]
= 40x4 + 8x3 - 40x - 5
[1]
Close
Example
f(x) = (x5 - 2x)(7x2 + 3)
f(1) = ?
Solution f(1) = ?
y = (x5 - 2x)(7x2 + 3)
y' = (5x4 - 2)(7x2 + 3) + (x5 - 2x)(7⋅2x1 + 0)
= (5x4 - 2)(7x2 + 3) + (x5 - 2x)⋅14x
f'(1) = (5⋅14 - 2)(7⋅12 + 3) + (15 - 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
y' = (5x4 - 2)(7x2 + 3) + (x5 - 2x)(7⋅2x1 + 0)
= (5x4 - 2)(7x2 + 3) + (x5 - 2x)⋅14x
f'(1) = (5⋅14 - 2)(7⋅12 + 3) + (15 - 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 = 1x3 + 2x
y' = ?
Solution y' = ?
y = 1x3 + 2x
y' = -3x2 + 2(x3 + 2x)2
y' = -3x2 + 2(x3 + 2x)2
Close
y = 1x3 + 2x
= (x3 + 2x)-1 - [1]
y' = -1⋅(x3 + 2x)-2⋅(3x2 + 2) - [2]
= -3x2 + 2(x3 + 2x)2
= (x3 + 2x)-1 - [1]
y' = -1⋅(x3 + 2x)-2⋅(3x2 + 2) - [2]
= -3x2 + 2(x3 + 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 = 4xx2 - 3
y' = ?
Solution y' = ?
y = 4xx2 - 3
y' = -4⋅(x2 - 3) - 4x⋅(2x1 - 0)(x2 - 3)2
= -4⋅(x2 - 3) - 4x⋅2x(x2 - 3)2
= -4x2 - 12 - 8x2(x2 - 3)2
= --4x2 - 12(x2 - 3)2
= 4x2 + 12(x2 - 3)2
y' = -4⋅(x2 - 3) - 4x⋅(2x1 - 0)(x2 - 3)2
= -4⋅(x2 - 3) - 4x⋅2x(x2 - 3)2
= -4x2 - 12 - 8x2(x2 - 3)2
= --4x2 - 12(x2 - 3)2
= 4x2 + 12(x2 - 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 = (2x2 - 1)8
y' = ?
Solution y' = ?
y = (2x2 - 1)8
y' = 8⋅(2x2 - 1)7⋅(2⋅2x1 - 0)
= 8(2x2 - 1)7⋅4x
= 32x(2x2 - 1)7
y' = 8⋅(2x2 - 1)7⋅(2⋅2x1 - 0)
= 8(2x2 - 1)7⋅4x
= 32x(2x2 - 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
x2 + y2 = 1
dydx = ?
Solution dydx = ?
x2 + y2 = 1
2x1 + 2y1y' = 0
2x + 2yy' = 0
x + yy' = 0
yy' = -x
y' = -xy
2x1 + 2y1y' = 0
2x + 2yy' = 0
x + yy' = 0
yy' = -x
y' = -xy
Close
Example
x3 + xy2 - 2y3 + 2 = 0
dydx at (1, 1)?
Solution dydx at (1, 1)?
x3 + xy2 - 2y3 + 2 = 0
3x2 + 1⋅y2 + x⋅2y1y' - 2⋅3y2y' + 0 = 0 - [1]
3x2 + y2 + 2xyy' - 6y2y' = 0
2xyy' - 6y2y' = -3x2 - y2
y'(2xy - 6y2) = -3x2 - y2
y' = -3x2 - y22xy - 6y2
[y'](x, y) = (1, 1) = -3⋅12 - 122⋅1⋅1 - 6⋅12
= -3⋅1 - 12 - 6⋅1
= -3 - 12 - 6
= -4-4
= 1
3x2 + 1⋅y2 + x⋅2y1y' - 2⋅3y2y' + 0 = 0 - [1]
3x2 + y2 + 2xyy' - 6y2y' = 0
2xyy' - 6y2y' = -3x2 - y2
y'(2xy - 6y2) = -3x2 - y2
y' = -3x2 - y22xy - 6y2
[y'](x, y) = (1, 1) = -3⋅12 - 122⋅1⋅1 - 6⋅12
= -3⋅1 - 12 - 6⋅1
= -3 - 12 - 6
= -4-4
= 1
[1]
Close
Derivative of a Parametric Function
Formula
dydx = dydt dxdt
Example
x = t3 - 2t
y = t2 + 1
dydx at t = 1?
Solution y = t2 + 1
dydx at t = 1?
x = t3 - 2t
y = t2 + 1
dxdt = 3t2 - 2
dydt = 2t1 + 0
= 2t
dydx = dydt dxdt
= 2t3t2 - 2
[dydx]t = 1 = 2⋅13⋅12 - 2
= 23⋅1 - 2
= 23 - 2
= 21
= 2
y = t2 + 1
dxdt = 3t2 - 2
dydt = 2t1 + 0
= 2t
dydx = dydt dxdt
= 2t3t2 - 2
[dydx]t = 1 = 2⋅13⋅12 - 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 = x5 - x + 8
dxdy = ?
Solution dxdy = ?
y = x5 - x + 8
dydx = 5x4 - 1 + 0
= 5x4 - 1
dxdy = 15x4 - 1
dydx = 5x4 - 1 + 0
= 5x4 - 1
dxdy = 15x4 - 1
Close
Example
y = x3 + 2
dxdy at y = 3?
Solution dxdy at y = 3?
y = x3 + 2
dydx = 3x2 + 0
= 3x2
dxdy = 13x2
3 = x3 + 2 - [1]
1 = x3
x3 = 1
x = 1
dxdy = 13⋅12
= 13
dydx = 3x2 + 0
= 3x2
dxdy = 13x2
3 = x3 + 2 - [1]
1 = x3
x3 = 1
x = 1
dxdy = 13⋅12
= 13
[1]
The goal is to find dx/dy at y = 3.
dx/dy = 1/3x2
→ Find the x value when y = 3.
→ Put y = 3 into y = x3 + 2.
dx/dy = 1/3x2
→ Find the x value when y = 3.
→ Put y = 3 into y = x3 + 2.
Close
Second Derivative
How to Write
f''(x), y'', d2ydx2, d2dx2f(x)
These are the ways to write the second derivative.To find the second derivative,
differentiate f(x) twice.
Inflection Point
Example
f(x) = x5 - 7x2 - 8x + 1
f''(x) = ?
Solution f''(x) = ?
f(x) = x5 - 7x2 - 8x + 1
f'(x) = 5x4 - 7⋅2x1 - 8 + 0
= 5x4 - 14x - 8
f''(x) = 5⋅4x3 - 14 - 0
= 20x3 - 14
f'(x) = 5x4 - 7⋅2x1 - 8 + 0
= 5x4 - 14x - 8
f''(x) = 5⋅4x3 - 14 - 0
= 20x3 - 14
Close