Find the partial derivative with respect to Pa of V = 1000M(Pg - Pa) / [Pg(Pv - Pa)(1 + (t - 20)b)], treating M, Pg, Pv, t and b as constants.

Step-by-step solution

1. 1

   Pull out everything free of $P_a$. Only two factors contain $P_a$, one in the numerator and one in the denominator. Define the constant

   $C=\frac{1000M}{P_g\bigl(1+(t-20)b\bigr)},$

   so that

   $V=C\cdot\frac{P_g-P_a}{P_v-P_a}.$

   Isolating $C$ first is what keeps the differentiation to a single quotient rule instead of a product-and-quotient tangle.

2. 2

   Apply the quotient rule to the remaining ratio. With $f=P_g-P_a$ and $g=P_v-P_a$, both derivatives are $-1$:

   $\frac{\partial}{\partial P_a}\frac{f}{g}=\frac{f'g-fg'}{g^{2}}=\frac{-(P_v-P_a)+(P_g-P_a)}{(P_v-P_a)^{2}}.$

3. 3

   Simplify the numerator. The two $P_a$ terms cancel:

   $-(P_v-P_a)+(P_g-P_a)=-P_v+P_a+P_g-P_a=P_g-P_v.$

   So the derivative of the ratio is $\dfrac{P_g-P_v}{(P_v-P_a)^{2}}$ — a constant numerator over a square. That cancellation is the reason the answer is so compact.

4. 4

   Restore the constant factor.

   $\frac{\partial V}{\partial P_a}=\frac{1000M\,(P_g-P_v)}{P_g\bigl(1+(t-20)b\bigr)\,(P_v-P_a)^{2}}.$

5. 5

   Sanity-check the sign and the special case. The denominator $(P_v-P_a)^{2}$ is positive, so the sign of $\partial V/\partial P_a$ is fixed once and for all by whether $P_g$ exceeds $P_v$ — it never changes as $P_a$ varies. And if $P_g=P_v$ the derivative is identically zero, which is right, because then $V=C$ is a constant independent of $P_a$.