Chain Rule ALL

Chain Rule ALL

12th Grade

4 Qs

quiz-placeholder

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Chain Rule ALL

Chain Rule ALL

Assessment

Quiz

Mathematics

12th Grade

Easy

Created by

Jennifer Smith

Used 1+ times

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4 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

 y=(x2+5)5x5+2y=\frac{\left(-x^2+5\right)^{-5}}{x^5+2}  

 dydx=5x(3x5+45x3)(x5+2)(x2+5)6\frac{dy}{dx}=\frac{5x\left(3x^5+4-5x^3\right)}{\left(x^5+2\right)\left(-x^2+5\right)^6}  

 dydx=5x(x2+5)4(3x5+45x3)\frac{dy}{dx}=5x\left(-x^2+5\right)^4\left(3x^5+4-5x^3\right)  

 dydx=5x(3x5+45x3)(x5+2)2(x2+5)6\frac{dy}{dx}=\frac{5x\left(3x^5+4-5x^3\right)}{\left(x^5+2\right)^2\left(-x^2+5\right)^6}  

 dydx=5x(3x5+45x3)(x2+5)6\frac{dy}{dx}=\frac{5x\left(3x^5+4-5x^3\right)}{\left(-x^2+5\right)^6}  

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 y=(2x4+1)13(5x+3)y=\left(-2x^4+1\right)^{\frac{1}{3}}\left(5x+3\right)  

 dydx=70x4+1524x33(2x4+1)23\frac{dy}{dx}=\frac{-70x^4+15-24x^3}{3\left(-2x^4+1\right)^{\frac{2}{3}}}  

 dydx=5(2x4+1)13\frac{dy}{dx}=5\left(-2x^4+1\right)^{\frac{1}{3}}  

 dydx=80x33(2x4+1)23\frac{dy}{dx}=-\frac{80x^3}{3\left(-2x^4+1\right)^{\frac{2}{3}}}  

 dydx=15(2x4+1)238x33(2x4+1)23\frac{dy}{dx}=\frac{15\left(-2x^4+1\right)^{\frac{2}{3}}-8x^3}{3\left(-2x^4+1\right)^{\frac{2}{3}}}  

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 y=(x5+3x23)14y=\left(\frac{x^5+3}{-x^2-3}\right)^{\frac{1}{4}}  

 dydx=3x(x55x3+2)(x23)2\frac{dy}{dx}=\frac{3x\left(-x^5-5x^3+2\right)}{\left(-x^2-3\right)^2}  

 dydx=3x(x55x3+2)4(x5+3)34(x23)54\frac{dy}{dx}=\frac{3x\left(-x^5-5x^3+2\right)}{4\left(x^5+3\right)^{\frac{3}{4}}\left(-x^2-3\right)^{\frac{5}{4}}}  

 dydx=3x(x55x3+2)(x5+3)34(x23)54\frac{dy}{dx}=\frac{3x\left(-x^5-5x^3+2\right)}{\left(x^5+3\right)^{\frac{3}{4}}\left(-x^2-3\right)^{\frac{5}{4}}}  

 dydx=(x23)344(x5+3)34\frac{dy}{dx}=\frac{\left(-x^2-3\right)^{\frac{3}{4}}}{4\left(x^5+3\right)^{\frac{3}{4}}}  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 y=(3x2+2)2(x33)y=\left(3x^2+2\right)^2\left(x^3-3\right)  

 dydx=3x(12x2+8+x)\frac{dy}{dx}=3x\left(12x^2+8+x\right)  

 dydx=72x3(3x2+2)\frac{dy}{dx}=72x^3\left(3x^2+2\right)  

 dydx=(3x2+2)2(3x2)\frac{dy}{dx}=\left(3x^2+2\right)^2\left(3x^2\right)  

 dydx=3x(3x2+2)(7x3+2x12)\frac{dy}{dx}=3x\left(3x^2+2\right)\left(7x^3+2x-12\right)  

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