Chain Rule : If f(x) and g(x) are both differentiable functions and we describe F(x) = (f. g)(x) so the derivative of F(x) is F′(x) = f ′(g(x)) g′(x).
Proof
We will start off the proof through defining u = g(x) and considering that in terms of this definition what we are going asked to prove,
(d/dx)[f(u)] = f'(u) (du/dx)
Let's consider the derivative of u(x) (again, keep in mind we have defined u(g(x)) and therefore u actually is a function of x) that we know exists since we are assuming that g(x) is differentiable. Through definition we contain,
u'(x) = limh→0 (u(x + h) - u(x)/h)
Remember as well that,
limh→0 ((u(x + h) - u(x)/h) - u'(x))
= limh→0 (u(x + h) - u(x)/h) - limh→0 u'(x)
= u'(x) - u'(x) = 0
Now, describe as,
If h ≠ 0 then v(h) = (u(x + h) - u(x))/h - u'(x)
If h = 0 then v(h) = 0
and remember that limh→0v(h) =0 = v(0) and therefore v(h) is continuous at h = 0.
Now if we suppose that h ≠ 0 we can rewrite the definition of v (h) to find,
u(x + h) = u(x) + h(v( h) + u′(x) (1)
Now, notice that (1) is in fact valid even if we let h ≠ 0 and so is valid for any value of h.
After that, since we also know that f(x) is differentiable we can do something same. Though, we're going to utilize a diverse set of letters or variables here for causes which will be apparent in a little.
Therefore define,
If k ≠ 0 then w(k) = (f(z + k) - f(z))/k - f'(x)
If k = 0 then w(k) = 0
We can suffer a same argument that we did above thus demonstrate that w(k) is continuous at k = 0 and that,
f(z + k) = f(z) + k(w(k) + f ′(z)) (2)
Do not get excited regarding the different letters at this point all we did was, by use k in place of h and let x = z. Nothing fancy at this time, although the change of letters will be helpful down the road.
Then, to this point this doesn't look like we have actually done anything hwihc gets us even close to giving the chain rule. The work above will produce very significant in our proof however therefore let's get going onto the proof.
What we require to do at this time is utilize the definition of the derivative and estimate the following limit.
(d/dx)[f[u(x)]] = limh→0 (f[u(x + h) - f(u(x))])/h (3)
Remember that even if the notation is more than a little messy if we utilize u(x) in place of u we require reminding ourselves here that u actually is a function of x.
Let's this time use (1) to rewrite the u(x+ h) and yes the notation is going to be unpleasant although we are going to have to deal along with this. By using (1), the numerator into the limit above turns into,
f[u(x + h)] - f[u(x)] = f[u(x) + h(v(h) + u'(x))] - f[u(x)]
If we then describe z = u(x) and k = h(v(h)) + u′(x)) we can utilize (2) to then write this as,
f[u(x + h)] - f[u(x)] = f[u(x) + h(v(h) + u'(x))] - f[u(x)]
= f[u(x)] + h(v(h) + u'(x))(w(k) + f'[u(x)]) - f[u(x)]
= h(v(h) + u'(x)) (w(k) + f'[u(x)])
Remember that we were capable to cancel a f[u(x)] to simplify all things up a bit. Also, remember that the w(k) was intentionally left that method to maintain the mess to a minimum now, just notice that k = h(v(h) + u′(x)) here as which will be significant at this point in a little. Let's here go back and keep in mind that all that was the numerator of our limit, (3). Plugging it in (3) provides,
d/dx)[f[u(x)]] = limh→0 (h(v(h) + u'(x)) (w(k) + f'[u(x)])
= limh→0 (v(h) + u'(x))(w(k) + f'[u(x)])
Notice that the h's canceled out. Next, recall that k = h (v(h) + u′ (x) and therefore,
= limh→0 k = lim h→0 (h(v(h) + u′(x)) = 0
Though, if limh→0 k = 0, as we have defined k anyway, so by the definition of w and the fact that we know w(k) is continuous at k = 0 we also know that,
limh→0 w(k) = w(limh→0 k)
= w(0)
= 0
And, recall that limh→0v(h) = 0. By using all of these facts our limit is,
d/dx)[f[u(x)]] = limh→0 (v(h) + u'(x)) (w(k) + f'[u(x)])
= u'(x) f'[u(x)]
= f'[u(x)] du/dx
It is exactly what we required to prove and therefore we're done.