Answer
The detailed explanations are below.
Work Step by Step
a) We examine $\frac{d}{dx}(\frac{1}{v})$:
According to the Derivative Quotient Rule, we have: $$\frac{d}{dx}\Big(\frac{1}{v}\Big)=\frac{v\frac{d}{dx}(1)-1\frac{dv}{dx}}{v^2}=\frac{0\times v-\frac{dv}{dx}}{v^2}=-\frac{\frac{dv}{dx}}{v^2}=-\frac{1}{v^2}\frac{dv}{dx}$$
This is in fact the Reciprocal Rule. So the Reciprocal Rule is in essence still the Derivative Quotient Rule for $\frac{d}{dx}(\frac{u}{v})$, yet now with $u=1$.
b) We can use the Derivative Product Rule together with the Reciprocal Rule to deduce the Derivative Quotient Rule, as follows:
$$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{d}{dx}\Big(u\times\frac{1}{v}\Big)$$
Apply the Derivative Product Rule here: $$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{du}{dx}\times\frac{1}{v}+\frac{d}{dx}\Big(\frac{1}{v}\Big)\times u$$
Next, we use Reciprocal Rule here to calculate $\frac{d}{dx}(\frac{1}{v})$:
$$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{du}{dx}\times\frac{1}{v}+\Big(-\frac{1}{v^2}\frac{dv}{dx}\Big)u$$
$$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{\frac{du}{dx}}{v}-\frac{u\frac{dv}{dx}}{v^2}$$
$$\frac{d}{dx}\Big(\frac{u}{v}\Big)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$
We have thus proved the Derivative Quotient Rule.