$x^{\ 2}+y^{\ 2}=r^{\ 2}$  
 $m_{\tan}$  at point  $P\left(x,y\right)$  
 $\frac{dy}{dx}=-\ \frac{A}{B}$  
Submit  $A,B$  
(small letters)

(a)  

 $x^{\ 2}+y^{\ 2}=2809$  
 $m_{\tan}$  at point  $P\left(-\ 45\ ,\ A\right)$ 
Point  $P$  lies in quadrant II 
 $\frac{dy}{dx}=\frac{B}{D}$  
Submit  $A,B,D$  

(a)  

 $x^{\ 10}-\ y^{\ 6}=16$  
 $A\ x^{\ B}dM-D\ y^{\ E}\ dN=G$  
 $A\ x^{\ B}dM=D\ y^{\ E}\ dN$  
 $\frac{H\ x^{\ B}}{K\ y^{\ E}}=\frac{dy}{dx}$  
Submit  $A,B,D,E,G,H,K,M,N$  

(a)  

A ball is dropped vertically downward from the top of a $157\ ft$  building.

 $s\left(t\right)=A\ t^{\ 2}+B\ t\ +D$  

(a)  

Position function
 $s\left(t\right)=7\ t^{\ 2}+9\ t\ +5$  

Find the velocity of the object at  $t=12$  

(a)  

 $y=96\ \cos\left(96\ x\right)$  
 $\frac{dy}{dx}=A\ trig\left(96\ x\right)$  
Submit  $A,trig$  

(a)  

 $y=1229\ \cot^{\ 25}x$  
 $\frac{dy}{dx}=A\ trig1^Bx\ trig2^{\ 2}x$  
Submit  $A,B,trig1,trig2$  

(a)  

 $y=45\ \sec\left(9\ x^{\ 5}\right)$  
 $\frac{dy}{dx}=$  
 $A\ x^{\ B}\ trig1\left(9\ x^3\right)trig2\left(9\ x^3\right)$  
Submit  $A,B,trig1,trig2$  

(a)  

Implicit Derivative
 $5\ x^7\left(3\ y^2dy\right)+y^3\left(35\ x^6dx\right)=0$  

General Solution ( $C$  is a constant )

(a)  

Implicit Derivative
 $\frac{x\ \sec^2y\ dy-\tan y\ dx}{x^2}=0$  

General Solution ( $C$  is a constant )

(a)