So, something like $\lim_{h\to 0}\frac{f(g(x+h))-f(g(x))}{h}$?
#$\dv{x}f(g(x))$ using definition of limits?
clever fable
upbeat monolithBOT
Kocher
inland stoneBOT
So, something like $\lim_{h\to 0}\frac{f(g(x+h))-f(g(x))}{h}$?
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carmine jolt
**Kocher**
Try multiplying and dividing this fraction by (g(x + h) - g(x)).
Then you just have two quotients: one for df/dg, other for dg/dx.
clever fable
**Kocher**
I also know this is the same as $\lim_{h\to 0}\frac{f(g(x)+h)-f(g(x))}{h}\cdot\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$
upbeat monolithBOT
Kocher
clever fable
**Kocher**
$\lim_{h\to 0}\frac{(f(g(x+h))-f(g(x)))(g(x+h)-g(x))}{h(g(x+h)-g(x))}$
upbeat monolithBOT
Kocher
carmine jolt
Now separate the two fractions.
clever fable
$\lim_{h\to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot g’(x)$
upbeat monolithBOT
Kocher
carmine jolt
Yeah.
clever fable
u=g(x+h)-g(x)?
carmine jolt
Well, sure, why not. Then the first fraction is (f(g + u) - f(g))/u.
clever fable
Well, sure, why not. Then the first fraction is (f(g + u) - f(g))/u.
How?
clever fable
If u = g(x + h) - g(x), then g(x + h) = u + g(x).
I get that, I don’t understand f(g).
carmine jolt
What do you mean? That's the second term in the numerator.
carmine jolt
No worries 😄
You're welcome!
clever fable
I never learned the derivations so I just want to reinforce it.
Wait one more question
carmine jolt
And I think you can see how this generalizes to any number of nested functions.
Yeah, what is it?
clever fable
$\int_a^b\cos(x)dx$ with definition of definite integral
upbeat monolithBOT
Kocher
clever fable
It just popped up in my mind, so you don’t have to answer it if you don’t want to.
carmine jolt
Oof...
I mean, we can try working through it, I guess.
It will involve a sum of cosines of an arithmetic progression, and I have a formula for that.
One sec. Let me share the formula, the rest I think you can try yourself, as I'm sure you won't have a problem writing the Riemann sum.
clever fable
Ok.
carmine jolt
Here you go.
clever fable
$\lim_{n\to\infty}\sum_{k=1}^n\frac{b-a}{n}\cos(a+\frac{b-a}{n}k)$
upbeat monolithBOT
Kocher
carmine jolt
What's wrong?
clever fable
Too many variable 
I do know that $\sum_{k=1}^n\cos(ck)=\Re\qty(\sum_{k=1}^n e^{ikc})$
carmine jolt
Well, let's look at your case.
a + (b - a)(k/n) = a + (b - a)((k - 1)/n + 1/n) = a + (b - a)/n + (b - a)((k - 1)/n)
So, in your case φ1 = a + (b - a)/n, Δφ = (b - a)/n.
upbeat monolithBOT
Kocher
clever fable
Wait, why is that complex?
carmine jolt
The formula I gave?
clever fable
Nevermind
carmine jolt
Well, its real part gives the sum of cosines and the imaginary part gives the sum of sines. So, kinda two in one.
obsidian runeBOT
@clever fable
tall hull
$\lim_{h\to 0}\frac{(f(g(x+h))-f(g(x)))(g(x+h)-g(x))}{h(g(x+h)-g(x))}$
I am late to the game but this is illegal
Because you don't know if g(x+h) - g(x) = 0 or not
The proof of the chain rule should not be in the hs program
But the idea is to use this characterization:
A continuous function f is differentiable at x, if and only if, there exists a number k and a function u such that for all h:
f(x+h) - f(x) = kh + u(h)
Where u(h)/h is known to tend to zero. In this case, k = f'(x)
Proof:
<= is trivial.
=>: Denote u(h) = f(x+h) - f(x) - f'(x)h and let k = f'(x). Just verify that u(h)/h tends to zero, and that the property is verified
You can use this characterization. If g is differentiable at x and f is differentiable at g(x), you want to prove that f o g is differentiable at x. Alternatively, you want to prove there exists a number k and a function u such that:
f(g(x+h)) - f(g(x)) = kh + u(h)
tall hull
Try multiplying and dividing this fraction by (g(x + h) - g(x)).
I think this comment will be helpful for you too
clever fable
I am late to the game but this is illegal
👍