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Calc: Quotient and Chain Rule Quiz

Authored by Corey Shepherd

Mathematics

11th - 12th Grade

CCSS covered

Used 6+ times

Calc: Quotient and Chain Rule Quiz
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22 questions

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1.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

What is NOT an acceptable way to write the quotient rule for the derivative of y =  fg\frac{f}{g} 

 y′=(gf′−fg′)g2y'=\frac{(gf'-fg')}{g^2}  

 y′=(f′g−g′f)g2y'=\frac{(f'g-g'f)}{g^2}  

 y′=lo⋅dhi−hi⋅dlolo2y'=\frac{lo\cdot dhi-hi\cdot dlo}{lo^2}  

 y′=(fg′−gf′)g2y'=\frac{(fg'-gf')}{g^2}  

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

True or False:  ddx((f(g(x))))=f′(g′(x))\frac{d}{dx}\left(\left(f\left(g\left(x\right)\right)\right)\right)=f'\left(g'\left(x\right)\right)  

True

False

Can't tell, it depends on the rule

It's all the parentheses for me

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which rule can you NOT use the chain rule on?

 (8x3+π)10\left(8x^3+π\right)^{10}  

 ln⁡(5x3−3)\ln\left(5x^3-3\right)  

 tan⁡(400x2)\tan\left(400x^2\right)  

 ex(sin⁡x+cos⁡x)e^x\left(\sin x+\cos x\right)  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which rule can you NOT use the quotient rule on?  \  

 cot⁡x\cot x  

 exsec⁡x\frac{e^x}{\sec x}  

 ln⁡(4x)\ln\left(4x\right)  

Chai x+66x2\frac{x+6}{6x^2}  n Rule

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which of the following doesn't require the quotient rule? Meaning it can be rewritten and differentiated with a different rule.

 sin⁡xx2\frac{\sin x}{x^2}  

 x4−5x+64x3+2x\frac{x^4-5x+6}{4x^3+2x}  

 ln⁡xex\frac{\ln x}{e^x}  

 5x4\frac{5}{x^4}  

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 y=csc⁡8xy=\csc^8x  Which of the following is the proper way to setup up the chain rule?   \  

 y=u8, u = csc⁡xy=u^8,\ u\ =\ \csc x  

 y=csc⁡u, u = x8y=\csc u,\ u\ =\ x^8  

Not a composite function

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Set up the derivative of y(x)=3x−7(x2+2x+1)y\left(x\right)=\frac{3x-7}{\left(x^2+2x+1\right)}  

 y′(x)=(3x−7)(2x+2)−(3)(x2+2x+1)(3x−7)2y'\left(x\right)=\frac{\left(3x-7\right)\left(2x+2\right)-\left(3\right)\left(x^2+2x+1\right)}{\left(3x-7\right)^2}  

 y′(x)=(3x−7)(2x+2)−3(x2+2x+1)(x2+2x+1)2y'\left(x\right)=\frac{\left(3x-7\right)\left(2x+2\right)-3\left(x^2+2x+1\right)}{\left(x^2+2x+1\right)^2}  

 y′(x)=(3)(x2+2x+1)−(3x−7)(2x+2)(x2+2x+1)y'\left(x\right)=\frac{\left(3\right)\left(x^2+2x+1\right)-\left(3x-7\right)\left(2x+2\right)}{\left(x^2+2x+1\right)}  

 y′(x)=(3)(x2+2x+1)−(3x−7)(2x+2)(x2+2x+1)2y'\left(x\right)=\frac{\left(3\right)\left(x^2+2x+1\right)-\left(3x-7\right)\left(2x+2\right)}{\left(x^2+2x+1\right)^2}  

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