How to Obtain This Partial Differential Equation Transformation?

Question

Given that $z=\frac{A_b}{B(a,b)}$ and $S(z,b)=\frac{P(a,A_b,b)}{B(a,b)}$, then we have the following formula:$\frac{\partial P}{\partial b}=S\frac{\partial B}{\partial b}+B\frac{\partial S}{\partial b}+B\frac{\partial S}{\partial z}\frac{\partial z}{\partial b}$. Can someone explain to me how they get that formula? I am desperate now. Thank you so much!

Answer 1

$$P=B(a,b)S(z,b)$$ You will need $\frac{\partial S}{\partial b}$. But there is a snare because $z$ is also function of $b$.

In order to escape to the trap, we define $S$ as a function of two variables $z$ and $y$ , that is $S(z,y)$ : $$S(a,b)=S\bigg(z(a,b), y(a,b)\bigg)$$

$$\frac{\partial S}{\partial b}= \frac{\partial S}{\partial z} \frac{\partial z}{\partial b}+ \frac{\partial S}{\partial y} \frac{\partial y}{\partial b}$$

In the present case $y$ is the particular function $y(a,b)=b$, thus $\frac{\partial y}{\partial b}=1$ and : $$\frac{\partial S}{\partial b}= \frac{\partial S}{\partial z} \frac{\partial z}{\partial b}+ \frac{\partial S}{\partial y}$$

The trap is made obvious : if we confuse $y$ with $b$ because $y(b)=b$ we will write $$ \color{red}{\frac{\partial S}{\partial b}}= \frac{\partial S}{\partial z} \frac{\partial z}{\partial b}+ \color{red}{\frac{\partial S}{\partial b}}\quad\text{which seems absurd.}$$ As a consequence we don't confuse $y(b)$ with $b$.

$$P=B(a,b)S(z,b)$$ $$\frac{\partial P}{\partial b}=S \frac{\partial B}{\partial b}+B \frac{\partial S}{\partial b}$$

$$\frac{\partial P}{\partial b}=S \frac{\partial B}{\partial b}+B \left(\frac{\partial S}{\partial z} \frac{\partial z}{\partial b}+ \frac{\partial S}{\partial y}\frac{\partial y(b)}{\partial b}\right)$$

$$\frac{\partial P}{\partial b}=S \frac{\partial B}{\partial b}+B\frac{\partial S}{\partial z} \frac{\partial z}{\partial b}+ B\frac{\partial S}{\partial y}\frac{\partial y(b)}{\partial b}$$

Of course these precautions are for people not fully familiar with the common way to deal with partial derivatives. Loosely one will commonly simplify the writing as : $$\boxed{\frac{\partial P}{\partial b}=S \frac{\partial B}{\partial b}+B\frac{\partial S}{\partial z} \frac{\partial z}{\partial b}+ B\frac{\partial S}{\partial b}}$$