Partial derivatives where implicit equations are involved

Question

I have a function: $$f\bigl(P,R\bigr)=h\bigl(P,R,N(P,R)\bigr)$$ where $P$ and $R$ are independent variables and $N$ is the solution of the equation: $$P-g(R,N)=0$$ I am trying to find $\frac{\partial f}{\partial P}$ and $\frac{\partial f}{\partial R}$.

Answer 1

$N(P,R)$ is not an independent variable, rather a function (of $P$ and $R$). Hence partial derivatives of $f$ with respect to $P$ and $R$ exist.

The solution is to obtain $\frac{\partial g}{\partial R}$ and $\frac{\partial g}{\partial N}$ from the solution of $P-g(R,N)=0$.

At the solution, P = g, and hence: $$\frac{\partial N}{\partial P}=\frac{1}{\frac{\partial g}{\partial N}}$$ $$\frac{\partial N}{\partial R}=-\frac{\frac{\partial g}{\partial R}}{\frac{\partial g}{\partial N}}$$ Use these derivatives with the chain rule in the evaluation of $$\frac{\partial}{\partial P}f\bigl(P,R\bigr)$$and $$\frac{\partial}{\partial R}f\bigl(P,R\bigr)$$

Answer 2

To obtain $\frac{\partial f}{\partial P}$, we compute $$\frac{\partial f}{\partial P}(P,R) = \frac{\partial h}{\partial P}(P,R,N(P,R)) + \frac{\partial h}{\partial N}(P,R,N(P,R))\cdot \frac{\partial N}{\partial P}(P,R).\tag{1}$$ Now, to compute $\frac{\partial N}{\partial P}$ we use the chain rule and the equation $P-g(R,N)=0$. Differentiating with respect to $P$: $$1 - \frac{\partial g}{\partial N}(R,N)\cdot \frac{\partial N}{\partial P}(P,R) = 0.$$ This gives us $$1 = \frac{\partial g}{\partial N}(R,N)\cdot \frac{\partial N}{\partial P}(P,R) \implies \frac{\partial N}{\partial P}(P,R) = \frac{1}{\frac{\partial g}{\partial N}(R,N)}.$$ In all, then, by $(1)$ we get $$\boxed{\frac{\partial f}{\partial P}(P,R) = \frac{\partial h}{\partial P}(P,R,N(P,R)) + \frac{\frac{\partial h}{\partial N}(P,R,N(P,R))}{\frac{\partial g}{\partial N}(R,N)}.}$$

For the second part, $\frac{\partial f}{\partial R}$, we compute $$\frac{\partial f}{\partial R}(P,R) = \frac{\partial h}{\partial R}(P,R,N(P,R)) + \frac{\partial h}{\partial N}(P,R,N(P,R))\cdot \frac{\partial N}{\partial R}(P,R).\tag{2}$$ Now, as before, we differentiate $P-g(R,N)=0$ with respect to $R$ to obtain $\frac{\partial N}{\partial R}$: $$-\frac{\partial g}{\partial R}(R,N) - \frac{\partial g}{\partial N}(R,N)\cdot \frac{\partial N}{\partial R}(P,R) = 0 \implies \frac{\partial N}{\partial R}(P,R) = -\frac{\frac{\partial g}{\partial R}(R,N)}{\frac{\partial g}{\partial N}(R,N)}.$$ In all, then, by $(2)$ we get $$\boxed{\frac{\partial f}{\partial R}(P,R) = \frac{\partial h}{\partial R}(P,R,N(P,R)) - \frac{\partial h}{\partial N}(P,R,N(P,R))\cdot \frac{\frac{\partial g}{\partial R}(R,N)}{\frac{\partial g}{\partial N}(R,N)}.}$$