Answer
See explanations.
Work Step by Step
a. Step 1. Recall the formula for the Derivative Quotient Rule as $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$ and letting $u=1$, we have $\frac{d}{dx}(\frac{1}{v})=\frac{v\frac{d1}{dx}-\frac{dv}{dx}}{v^2}
=\frac{0-\frac{dv}{dx}}{v^2}=-\frac{1}{v^2}\frac{dv}{dx}=$
Step 2. Comparing the above result with the formula given in the Exercise, we can conclude that the Reciprocal Rule is a special case of the Derivative Quotient Rule when $u=1$
b. Step 1. Recall the formula for the Derivative Product Rule as: $\frac{d}{dx}(uw)=u\frac{dw}{dx}+w\frac{du}{dx}$
Step 2. Let $w=1/v$, use the Reciprocal Rule, we have $\frac{dw}{dx}=\frac{d}{dx}(\frac{1}{v})=-\frac{1}{v^2}\frac{dv}{dx}$
Step 3. The Derivative Product in step-1 becomes: $\frac{d}{dx}(\frac{u}{v})=u(-\frac{1}{v^2}\frac{dv}{dx})+\frac{1}{v}\frac{du}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$ which is the Derivative Quotient Rule.
