Search Header Logo

derivatives product quotient chain rule

Authored by Kathryn Morse

12th Grade

CCSS covered

Used 2+ times

derivatives product quotient chain rule
AI

AI Actions

Add similar questions

Adjust reading levels

Convert to real-world scenario

Translate activity

More...

    Content View

    Student View

13 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the derivative of xn?

(n-1)xn
nxn+1
(n+1)xn-1
nxn-1

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

find 

y = 2 x-1
y=2 x-2
y= −2x-2
y= −2 x-1

Tags

CCSS.HSF-IF.C.7A

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Set up the derivative of y=(3x- 7)*(5x+ 1)

(12x3)*(10x)
(3x4 - 7)*(10x) + (12x3)*(5x2 + 1)
(3x4 - 7)*(10x) - (12x3)*(5x2 + 1)
(3x4 - 7)/(10x) + (12x3)/(5x2 + 1)

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the derivative of sin(x)?

-sin(x)
-cos(x)
cos(x)
sin(x)

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Which of the following is the derivative of g(x)= (x2)(2x4 + 2)5 ?

g'(x)= (2x)(2x4 + 2) 5 + (x2)(5(2x4 + 2)4(8x3))

g'(x)= 2x + 8x3

g'(x)= (2x)(2x4 + 2)5 + (x2)(8x3)

g'(x)= (2x)(5(2x4 + 2)5(8x3)) + (x2)(2x4 + 2)

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Media Image

Which of the following is the quotient rule for derivatives?

h'(x)=((g(x)f'(x) + f(x)g'(x)) / (g(x))2

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

h'(x)=((g(x)f'(x) - f(x)g'(x)) / (g(x))2

h'(x)=((g'(x)f'(x) + f(x)g(x)) / (g(x))

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Which of the following is the derivative of the function h(x)= (8x6 + 2x + 5)4 ?

h'(x)= 4(8x6 + 2x + 5)3(48x5 + 2)

h'(x)= 4(48x5 + 2)3

h'(x)= 4(48x5 + 2 + 5)3

h'(x)= 3(8x6 + 2x + 5)4(48x5 + 2)2

Access all questions and much more by creating a free account

Create resources

Host any resource

Get auto-graded reports

Google

Continue with Google

Email

Continue with Email

Classlink

Continue with Classlink

Clever

Continue with Clever

or continue with

Microsoft

Microsoft

Apple

Apple

Others

Others

Already have an account?