substitution of dx with 0

A

AbdelRahmanShady

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Kind of stupid question, but I am self learning calculus and is trying to prove rules along way like chain rule and power rule. My question is that I found that I cant substitute some dxs in equation with zero and leave others or I get wrong answer. But if do the substitution at one go, answer is right. so am I right or there are cases where even my second method fails. One more thing I sometimes use f(x)dx in other function as limit to zero as f(x)dx will approach zero with time. I used this fact mainly when proving composition rule but is this method correct or I just lucked out.
 
AbdelRahmanShady said:
Kind of stupid question, but I am self learning calculus and is trying to prove rules along way like chain rule and power rule. My question is that I found that I cant substitute some dxs in equation with zero and leave others or I get wrong answer. But if do the substitution at one go, answer is right. so am I right or there are cases where even my second method fails. One more thing I sometimes use f(x)dx in other function as limit to zero as f(x)dx will approach zero with time. I used this fact mainly when proving composition rule but is this method correct or I just lucked out.
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I can't think of any case where you would do the substitution dx = 0. Can you show us an example?

-Dan
 
AbdelRahmanShady said:
Kind of stupid question, but I am self learning calculus and is trying to prove rules along way like chain rule and power rule. My question is that I found that I cant substitute some dxs in equation with zero and leave others or I get wrong answer. But if do the substitution at one go, answer is right. so am I right or there are cases where even my second method fails. One more thing I sometimes use f(x)dx in other function as limit to zero as f(x)dx will approach zero with time. I used this fact mainly when proving composition rule but is this method correct or I just lucked out.
Click to expand...
You mention proofs; I suspect what you mean is that you are using the definition of the derivative, where you take the limit as [imath]\Delta x[/imath] approaches zero. Is that what you mean by "dx"?

If so, then you never just replace [imath]\Delta x[/imath] with zero; you are taking a limit, which can sometimes be done by such a substitution. But in doing so, you would never do it in only one place at a time.

But we can give a much better answer if you show an example where you did something that didn't work.
 
something like proof of chain rule
f(x) * f(y); derivative = [math]\frac{f(x + dx) * g(x + dx) - f(x)g(x)} {dx}[/math]= [math]\frac{ (f(x)+ f'(x)dx) * (g(x) + g'(x)dx) - f(x)g(x)} {dx}[/math][math]\frac{ f(x)g(x) + f(x)g'(x)dx + g(x)f'(x)dx + f'(x)g'(x)dx^2 - f(x)g(x)} {dx}[/math]= [math]f(x)g'(x)+ g(x)f'(x)+ f'(x)g'(x)dx[/math]taking limit = [math]f(x)g'(x)+ g(x)f'(x) [/math] this is what i call substitution

but if i removed dx and replaced it by zero in upper fraction in first equation only I will not get write answer
 
AbdelRahmanShady said:
something like proof of chain rule
f(x) * f(y); derivative = [math]\frac{f(x + dx) * g(x + dx) - f(x)g(x)} {dx}[/math]= [math]\frac{ (f(x)+ f'(x)dx) * (g(x) + g'(x)dx) - f(x)g(x)} {dx}[/math][math]\frac{ f(x)g(x) + f(x)g'(x)dx + g(x)f'(x)dx + f'(x)g'(x)dx^2 - f(x)g(x)} {dx}[/math]= [math]f(x)g'(x)+ g(x)f'(x)+ f'(x)g'(x)dx[/math]taking limit = [math]f(x)g'(x)+ g(x)f'(x) [/math] this is what i call substitution

but if i removed dx and replaced it by zero in upper fraction in first equation only I will not get write answer
Click to expand...
This is wrong because you never mention limits! You must maintain, throughout your work, the fact that you are taking a limit, until, at the end, you do so. And when you do that, it should be clear where, in substituting, you would not be taking a limit correctly.
 
no sure what you mean. if you mean I shoud mention dx approaches 0 in beginning then ok sorry forgot to mention
 
AbdelRahmanShady said:
something like proof of chain rule
f(x) * f(y); derivative = [math]\frac{f(x + dx) * g(x + dx) - f(x)g(x)} {dx}[/math]= [math]\frac{ (f(x)+ f'(x)dx) * (g(x) + g'(x)dx) - f(x)g(x)} {dx}[/math][math]\frac{ f(x)g(x) + f(x)g'(x)dx + g(x)f'(x)dx + f'(x)g'(x)dx^2 - f(x)g(x)} {dx}[/math]= [math]f(x)g'(x)+ g(x)f'(x)+ f'(x)g'(x)dx[/math]taking limit = [math]f(x)g'(x)+ g(x)f'(x) [/math] this is what i call substitution

but if i removed dx and replaced it by zero in upper fraction in first equation only I will not get write answer
Click to expand...
What Dr.Peterson means is leave out the dx and replace it with [imath]\Delta x[/imath] and take a limit.

ie. [math]f(x)' = \lim_{ \Delta x \to 0} \dfrac{f(x + \Delta x) - f(x)}{\Delta x}[/math]
dx is a specific kind of number, called an "infinitesimal" which is very very very small, but never 0.

-Dan
 
ok if i replace every dx with delta x is proof right. and Is my theory about substitution right. or there are hidden caveats
 
AbdelRahmanShady said:
ok if i replace every dx with delta x is proof right. and Is my theory about substitution right. or there are hidden caveats
Click to expand...
Write it all out, using limits everywhere, and then justify every step in terms of limits. Substitution can be used in evaluating limits, but only under certain circumstances. If you think of it all as mere substitution, then you will make mistakes. This is where the "caveats" hide!

It is not primarily a matter of whether you call it dx or delta x, but whether you think in terms of limits.
 
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