chain rule and implicit differentiation

 $\frac{5}{2\sqrt{5x-3}}$  

 $\frac{1}{2\sqrt{5x-3}}$  

 $\frac{5}{\sqrt{5x-3}}$  

 $-\sqrt{5x-3}$  

 $3x^2\ +3y^2\ =\ 6y$  

 $3x^2+3y^2y'\ =\ 3y^2\ +6xyy'$  

 $x^3+3y^2y'\ =\ 6yy'$  

 $3x^2\ +\ 3y^2y'\ =\ 3y^2\ +\ 3xyy'$  

 $\left(\cos\ \theta\right)\ \cdot\frac{d\theta}{dt}$  

 $\left(-\cos\ \theta\right)\ \cdot\frac{d\theta}{dt}$  

 $\sin\left(\frac{d\theta}{dt}\right)$  

all of these

$$-\frac{x}{y}$$  

$$-\frac{y}{x}$$  

$$-\frac{6y}{x}$$  

$$\frac{y}{x}$$  

 $\frac{dy}{dx}=\frac{-6x+5}{12y-7}$  

 $\frac{dy}{dx}=\frac{-6x+5}{-12y+7}$  

 $\frac{dy}{dx}=\frac{6x-5}{12y-7}$  

 $\frac{dy}{dx}=\frac{-3x+5}{6y-7}$  

 $\frac{dy}{dx}=\frac{3x-5}{6y-7}$