Partial derivative with respect to a function of variables

Question

I got the following equation, and want to solve the effect of $\Delta\alpha,\Delta\beta, and \Delta\gamma$ :

$g=\alpha\frac{d}{d+e+f}+\beta\frac{e}{d+e+f}+\gamma\frac{f}{d+e+f}$

However, $\alpha,\beta, and \gamma$ are all the functions, like the following:

$\alpha = \frac{a}{d}$ ; $\beta=\frac{b}{e}$ ; $\gamma=\frac{c}{f}$

So, the equation end up:

$g=\frac{a}{d}\frac{d}{d+e+f}+\frac{b}{e}\frac{e}{d+e+f}+\frac{c}{f}\frac{f}{d+e+f}$

How can I find the effect of Δα,Δβ,andΔγ when they are all in functions?

Answer 1

Let apply chain rule

we have

$$g=g(\alpha,\beta,\gamma) \quad g=\frac{a+b+c}{d+e+f}$$ and $$\alpha=\alpha(a,d) \quad \beta=\beta(b,e) \quad \gamma=\gamma(c,f)$$

thus

$$dg=\frac{\partial g}{\partial \alpha}d\alpha+\frac{\partial g}{\partial \beta}d\beta+\frac{\partial g}{\partial \gamma}d\gamma$$

with

$$\frac{\partial g}{\partial \alpha}=\frac{\partial g}{\partial a}\frac{\partial a}{\partial \alpha}+\frac{\partial g}{\partial d}\frac{\partial d}{\partial \alpha}=\frac{d}{d+e+f}+\frac{-1}{(d+e+f)^2}\cdot \left(-\frac{a}{\alpha^2}\right)=$$ $$=\frac{d}{d+e+f}+\frac{d^2}{a(d+e+f)^2}$$

and so on

$$\frac{\partial g}{\partial \beta}=\frac{\partial g}{\partial b}\frac{\partial b}{\partial \beta}+\frac{\partial g}{\partial e}\frac{\partial e}{\partial \beta}=...$$

$$\frac{\partial g}{\partial \gamma}=\frac{\partial g}{\partial c}\frac{\partial c}{\partial \gamma}+\frac{\partial g}{\partial f}\frac{\partial f}{\partial \gamma}=...$$