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November 8th, 2018, 03:12 PM   #11
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 Originally Posted by v8archie You aren't using the function he gave. His idea about $f'(0)$ is perfectly valid, although we'd normally have $h$ rather than $x$.
What idea? In what respect is the function he gave in his response to me not the function I was discussing. I am now very confused. November 8th, 2018, 04:03 PM   #12
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His function had no denominator. You took the difference quotient at $x=0$ (written with $x$ instead of $h$) to be the function. The actual function is $$f(x)=\sqrt{x}\sin{(x)}$$
The product rule depends on the existence of both functions and both derivatives. This is implied by the appearance of both in the formula. The formula is obtained by manipulations according to the theory of limits.
$$\frac{f(x+h)g(x+h)-f(x)g(x)}{h} = g(x)\frac{f(x+h)-f(x)}{h}+f(x+h)\frac{g(x+h)-g(x)}{h}$$
by adding and subtracting $g(x)f(x+h)$ in the numerator. Taking limits we get \begin{align}\frac{\mathrm d}{\mathrm dx}\big(f(x)g(x)\big) = \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} &= \lim_{h \to 0} g(x)\frac{f(x+h)-f(x)}{h}+f(x+h)\frac{g(x+h)-g(x)}{h} \\
&= \lim_{h \to 0} g(x) \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}+\lim_{h \to 0} f(x+h) \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} = g(x)f'(x)+f(x)g'(x)\end{align}
But this final step (splitting the single limit into many) is only valid if all those limits exist (and are finite).

A good book (e.g. Apostol - Theorem 4.1 in section 4.5) will state something like
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 Let $f$ and $g$ be two functions defined on a common interval. At each point where $f$ and $g$ have a derivative, the same is true of the sum $f+g$, the difference $f-g$, [and] the product $f \cdot g$ (my emphasis)
This doesn't say that the derivative of the product doesn't exist where the individual derivatives don't exist, only that we can apply the formulas where they do. November 8th, 2018, 05:06 PM   #13
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 Originally Posted by v8archie His function had no denominator. You took the difference quotient at $x=0$ (written with $x$ instead of $h$) to be the function.
Yes. That is where I got confused. Thanks. November 9th, 2018, 04:24 AM   #14
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 Originally Posted by Integraluser Given that $\displaystyle f(x) = f(x+h) - 5x^3 h$, find $\displaystyle f'(x)$? So our teacher solved it in two ways, one of them was this: . . . $\displaystyle f'(x) = 5x^3$
What makes you think that any differentiable function $f(x)$ exists such that $f(x) = f(x+h) - 5x^3 h$? November 9th, 2018, 06:57 AM   #15
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 Originally Posted by skipjack What makes you think that any differentiable function $f(x)$ exists such that $f(x) = f(x+h) - 5x^3 h$?
His proof could be the start of a proof by contradiction! Assume it exists, then $f'(x) = ...$. November 9th, 2018, 09:22 AM   #16
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 Originally Posted by skipjack What makes you think that any differentiable function $f(x)$ exists such that $f(x) = f(x+h) - 5x^3 h$?
I'd say that the word "given" is specifically intended to bypass that question. November 9th, 2018, 10:49 AM   #17
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 Originally Posted by skipjack What makes you think that any differentiable function $f(x)$ exists such that $f(x) = f(x+h) - 5x^3 h$?
I really doubt it too but it's one among of my teacher's booklet questions or maybe the math book questions.
Just like archie mentioned, despite the fact that it's potentially not logical, it's intention is to bypass the question after all.

@Jeff, v8archie so apparently the difference is that limit of each function's slope has to exist, otherwise function wouldn't be differentiable using derivative rules. However, because formal definition combines both functions in a one function in way that would make it being treated as one entity which has its slope limit existing. Correct me if I'm wrong. November 9th, 2018, 11:11 AM   #18
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 Originally Posted by Integraluser @Jeff, v8archie so apparently the difference is that limit of each function's slope has to exist, otherwise function wouldn't be differentiable using derivative rules. However, because formal definition combines both functions in a one function in way that would make it being treated as one entity which has its slope limit existing. Correct me if I'm wrong.
More or less. By going from the formal definition, you are free to calculate the limit in whatever way works, without prior assumptions. By using the product rule, you are accepting that the limit has already been calculated based on the assumption that the functions and derivatives exist - you just plug in the expressions. November 9th, 2018, 01:09 PM   #19
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 Originally Posted by Integraluser Hey. So we had this question: Given that $\displaystyle f(x) = f(x+h) - 5x^3 h$, find $\displaystyle f'(x)$? So our teacher solved it in two ways, one of them was this: $\displaystyle f(x+h) - f(x) = 5x^3 h$ $\displaystyle \large\lim\limits_{h \to 0} \large\frac{f(x+h) - f(x)}{h} = \large\lim\limits_{h \to 0} \large\frac{5x^3 h}{h}$ $\displaystyle f'(x) = 5x^3$ My question is regarding the second step; how was he able to insert that limit? Did he multiply by the limit of (1/h) or divide by h then taken the limit of both sides or what? My second question is, what's the difference between first derivative formal definition and derivative rules regarding the way the work? And why can derivative rules have flaws in some points meanwhile first derivative formal definition doesn't have it at the same points?
Check the single variable OCW at the MIT webpage. There is a lecture (I don't remember which one) where the teacher get different derivation rules using the definition.

BR November 9th, 2018, 03:56 PM #20 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra I presume you mean this one: https://ocw.mit.edu/courses/mathemat...fferentiation/ which derives the normal rules from the limit (as I sketched above for the product rule). Tags derivatives, limits, questions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jkh1919 Calculus 9 October 6th, 2012 04:21 PM r-soy Calculus 3 November 28th, 2010 12:31 PM r-soy Calculus 3 November 4th, 2010 03:42 PM r-soy Calculus 2 November 3rd, 2010 11:31 PM r-soy Calculus 2 November 3rd, 2010 10:25 PM

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