Calc: Quotient and Chain Rule Quiz

 $y'=\frac{(gf'-fg')}{g^2}$  

 $y'=\frac{(f'g-g'f)}{g^2}$  

 $y'=\frac{lo\cdot dhi-hi\cdot dlo}{lo^2}$  

 $y'=\frac{(fg'-gf')}{g^2}$  

True

False

Can't tell, it depends on the rule

It's all the parentheses for me

 $\left(8x^3+π\right)^{10}$  

 $\ln\left(5x^3-3\right)$  

 $\tan\left(400x^2\right)$  

 $e^x\left(\sin x+\cos x\right)$  

 $\cot x$  

 $\frac{e^x}{\sec x}$  

 $\ln\left(4x\right)$  

Chai $\frac{x+6}{6x^2}$  n Rule

 $\frac{\sin x}{x^2}$  

 $\frac{x^4-5x+6}{4x^3+2x}$  

 $\frac{\ln x}{e^x}$  

 $\frac{5}{x^4}$  

 $y=u^8,\ u\ =\ \csc x$  

 $y=\csc u,\ u\ =\ x^8$  

Not a composite function

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

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

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

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

 $y'=\frac{8x^2+8x+5}{\left(x+3\right)^2}$ 

 $y'=\frac{4x^2+24x+5}{\left(x+3\right)^2}$ 

 $y'=\frac{4x^2+24x-5}{\left(x+3\right)^2}$ 

 $y'=\frac{8x^2+24x+5}{\left(4x^2-5\right)^2}$ 

 $y'\left(x\right)=\frac{x\tan x+4}{\sec x}$  

 $y'\left(x\right)=\frac{2x^3\left(4-x\tan x\right)}{\sec x}$  

 $y'\left(x\right)=\frac{2x^3\sec x\left(4-x\tan x\right)}{\left(2x^4\right)^2}$  

 $y'\left(x\right)=\frac{8x^3\sec x-2x^4\sec x\tan x}{\sec^2x}$  

 $y'\left(x\right)=\frac{-\cos x\cdot e^x+\sin x\cdot e^x}{\left(e^x\right)^2}$  

 $y'\left(x\right)=\frac{\cos x-\sin x}{e^x}$  

 $y'\left(x\right)=\frac{-\cos x+\sin x}{e^x}$  

 $y'\left(x\right)=\frac{-\sin x\cos x-e^{2x}}{\left(-\sin x^2\right)}$