Chain rule, Mathematics

Assignment Help:

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.


Related Discussions:- Chain rule

Proportional Relationships, Carmen bought 3 pounds of bananas for $1.08. Ju...

Carmen bought 3 pounds of bananas for $1.08. June paid for her purchase of bananas. If they paid the same price per pound, how many pounds did June buy?

Simple interest, find the simple interest on Rs. 68,000 at 50/3 per annum f...

find the simple interest on Rs. 68,000 at 50/3 per annum for 9 month

Basic mathematics, I need help with my homework, I am to the edge right now...

I need help with my homework, I am to the edge right now with this w=5pq/2

Regression, regression line drawn as Y=C+1075x, when x was 2, and y was 239...

regression line drawn as Y=C+1075x, when x was 2, and y was 239, given that y intercept was 11. calculate the residual

External division of section formula, give me the derivation of external di...

give me the derivation of external division of sectional formula using vectors

Math, is this free for LIFE that means forever never ever going to pay

is this free for LIFE that means forever never ever going to pay

Fractions, what is the lowest term of 11/121

what is the lowest term of 11/121

H, 6987+746-212*7665

6987+746-212*7665

Calculate the gross pay, 1. Simon's monthly take home pay (after taxes) is ...

1. Simon's monthly take home pay (after taxes) is $2200, if he pays 19%  of his gross pay(before taxex) in tax, what is his gross pay? 2 . Convert the following quantities to th

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd