Graphs of Derivatives

first derivative

second derivative

third derivative

expression

4

x

-4

0

$$\frac{dy}{dx}=18x^2$$  

$$\frac{dy}{dx}=6x^2$$

$$\frac{dy}{dx}=18x$$  

6 

$$\frac{\text{d}y}{\text{d}x}=18x^2-6x+1$$  

$$\frac{\text{d}y}{\text{d}x}=18x^3-6x-7$$  

$$\frac{\text{d}y}{\text{d}x}=6x^2-3x+1$$  

$$\frac{\text{d}y}{\text{d}x}=18x^2-6x+1-7$$  

0

1

x

7x

$$\frac{1}{2\sqrt{x}}$$  

0

1

$$\frac{1}{x^2}$$  

$$4x^{-\frac{1}{3}}$$

$$6x^{-\frac{2}{3}}$$

$$6x^{-\frac{1}{3}}$$

$$4x^{\frac{1}{3}}$$

1 - 2x

-2x

-2

2x

$$\frac{dy}{dx}=6x+5$$  

$$\frac{dy}{dx}=x^3+\frac{5}{2}x^2-3x$$  

$$\frac{dy}{dx}=3x+5$$  

$$\frac{dy}{dx}=x-3$$  

$$6x^2-8x+7-5$$  

$$6x^2-8x+7$$  

$$6x^3-8x^2+7x-5$$  

$$2x^2-4x+7$$  

35(x + 4)⁵

35(x + 4)⁴

140(x + 4)⁵

140(x + 4)⁴

$$10\left(2x-5\right)^5$$  

$$12\left(2x-5\right)^5$$  

$$6\left(2x-5\right)^5$$  

$$24\left(2x-5\right)^5$$  

$$5x\left(2x^2-3x+1\right)^4$$  

$$\left(4x-3\right)\left(2x^2-3x+1\right)^4$$  

$$\left(16x-3\right)\left(2x^2-3x+1\right)^4$$   

$$\left(20x-15\right)\left(2x^2-3x+1\right)^4$$  

6x⁵ - 12x³+8x

6x⁵ - 16x³+8x

x⁶-4x⁴+4x²

6x⁵ - 16x^3-8x

slope of the tangent line

slope of the normal line

slope of the secant line

the function’s value

at C only

at points A, C and E only

at point B and D only

at points A and E only

at C only

at points A, C and E only

at point B and D only

at points A and E only

4

1

2

3

1

4

8

16

f'(x)

dy/dx

y'

slope of the tangent line

instantaneous rate of change

f'(x)

y'

dy

dy/dx

The function is increasing

The function is decreasing

only 3

only 2

only 1

until you can no longer derive it