The chain rule in multi-variable calculus: Generalized

by ayushkhaitan3437

Now we’ll discuss the chain rule for n-nested functions. For example, an n-nested function would be g=f_1(f_2(\dots(f_n(t))\dots). What would \frac{\partial g}{\partial t} be?

We know that

g(t+h)-g(t)=\frac{f_1(f_2(\dots(f_n(t+h))\dots)-f_1(f_2(\dots(f_n(t))\dots)}{f_2(\dots(f_n(t+h))\dots)-f_2(\dots(f_n(t))\dots)}.f_2(\dots(f_n(t+h))\dots)-f_2(\dots(f_n(t))\dots).

If f_2 is continuous, then

g(t+h)-g(t)=\frac{\partial f_1}{\partial f_2}.f_2(\dots(f_n(t+h))\dots)-f_2(\dots(f_n(t))\dots)+g_1 such that \lim_{[f_2(\dots(f_n(t+h))\dots)-f_2(\dots(f_n(t))\dots)]\to 0}g_1=0, which is equivalent to saying \lim\limits_{t\to 0}g_1=0.

In turn

f_2(\dots(f_n(t+h))\dots)-f_2(\dots(f_n(t))\dots)=\frac{\partial f_2}{\partial f_3}.f_3(\dots(f_n(t+h))\dots)-f_3(\dots(f_n(t))\dots)+g_2

such that \lim\limits_{t\to 0}g_2=0.

Hence, we have

g(t+h)-g(t)=\frac{\partial f_1}{\partial f_2}.(\frac{\partial f_2}{\partial f_3}.\left[f_3(\dots(f_n(t+h))\dots)-f_3(\dots(f_n(t))\dots)\right]+g_2)+g_1

Continuing like this, we get the formula

g(t+h)-g(t)=\frac{\partial f_1}{\partial f_2}.(\frac{\partial f_2}{\partial f_3}.(\dots(\frac{\partial f_n}{\partial t}.t+g_n)+g_{n-1})\dots)+g_2)+g_1

such that \lim\limits_{t\to 0}g_i=0 for all i\in \{1,2,3,\dots,n\}.

From the above formula, we get

\lim\limits_{t\to 0}g(t+h)-g(t)=\frac{\partial f_1}{\partial f_2}.\frac{\partial f_2}{\partial f_3}.\dots\frac{\partial f_n}{\partial t}.t

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