November 8th, 2018, 02:12 PM  #11 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  
November 8th, 2018, 03:03 PM  #12  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 Math Focus: Mainly analysis and algebra 
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 $$f(x)=\sqrt[3]{x}\sin{(x)}$$ 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 Quote:
 
November 8th, 2018, 04:06 PM  #13 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  
November 9th, 2018, 03:24 AM  #14 
Global Moderator Joined: Dec 2006 Posts: 20,390 Thanks: 2015  What makes you think that any differentiable function $f(x)$ exists such that $f(x) = f(x+h)  5x^3 h$?

November 9th, 2018, 05:57 AM  #15 
Senior Member Joined: Oct 2009 Posts: 753 Thanks: 261  
November 9th, 2018, 08:22 AM  #16 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 Math Focus: Mainly analysis and algebra  
November 9th, 2018, 09:49 AM  #17  
Newbie Joined: Jun 2018 From: Jordan Posts: 24 Thanks: 0  Quote:
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, 10:11 AM  #18  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 Math Focus: Mainly analysis and algebra  Quote:
 
November 9th, 2018, 12:09 PM  #19  
Newbie Joined: Apr 2015 From: Lima Posts: 17 Thanks: 1  Quote:
BR  
November 9th, 2018, 02:56 PM  #20 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 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).


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