If \(\large f(x)=x^n\text{ then }f'(x)=n\cdot x^{n-1}\) is often called the basic power rule.And also, what is this equation called?
I know the limit definition I'm 9th grader but I'll try to understandIf \(\large f(x)=x^n\text{ then }f'(x)=n\cdot x^{n-1}\) is often called the basic power rule.
To stabatpatriae: The fact you asked causes one to wonder if you are in a class studying calculus?
Do you know the limit definition of derivative? \(\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} = f'(x)\) if it exists??
Please tell us about your situation.
Do you know the binomial expansion of (a+b)^n?And also, what is this equation called?
To understand this, we need to know that for a positive integer \(n\)I know the limit definition I'm 9th grader but I'll try to understand
Do you know the binomial expansion of (a+b)^n?
I do but no need to explain, I understood it. ThanksDo you know the binomial expansion of (a+b)^n?
Thanks a lot, yes calculus is easy when you know algebraTo understand this, we need to know that for a positive integer \(n\)
\((x+h)^n=\sum\limits_{k = 0}^n {\dbinom{n}{k}{x^{n - k}}{h^k}} \)
thus \((x+h)^n=\sum\limits_{k = 0}^n {\dbinom{n}{k}{x^{n - k}}{h^k}}-f(x)=\sum\limits_{k = 1}^n {\dbinom{n}{k}{x^{n - k}}{h^k}} \)
Note that each term contains an \(h\) so dividing by \(h\) and finding the limit as \(h\to 0\) all that is left is \(n\cdot x^{n-1}\)
I hope you see that calculus is rather easy, but you need the algebra to get there.
Sorry, but I do not entirely buy your proof. You must rigorously show that g(x+h) goes to g(x)!Of course, that requires that you first have proved the "product rule": (fg)'= f'g+ fg'.
If F(x)= f(x)g(x) then F(x+ h)= f(x+h)g(x+ h) so F(x+h)- F(x)= f(x+h)g(x+h)- f(x)g(x)= f(x+h)g(x+h)- f(x)g(x+h)+ f(x)g(x+h)- f(x)g(x)
= (f(x+h)- f(x))g(x+ h)- f(x)(g(x+h)- g(x).
\(\displaystyle \frac{F(x+h)- F(x)}{h}= \frac{f(x+h)- f(x)}{h}g(x+h)+ f(x)\frac{g(x+h)- g(x)}{h}\).
Now, as h goes to 0 the two "difference quotients" go to the derivative and g(x+h) goes to g(x).
Yes, I did not explicitly state that f and g must be differentiable which would imply that they are continuous.Sorry, but I do not entirely buy your proof. You must rigorously show that g(x+h) goes to g(x)!
In my opinion just before proving the product rule you prove the lemma showing that g(x+h) goes to g(x). Then in the proof of the product rule you state that you are using that lemma. This is how I was taught the product rule and how I taught it for decades. If I recall correctly, it is that way in all the calculus books I looked at. If it wasn't like that in the textbook I would never had said anything to Halls about it.This is one of the reasons that this site is not (and cannot be) well adapted for proofs: what theorems may be assumed as already proven and what axioms apply will vary with each inquiry.
[MATH]F(x) = f(x) \cdot g(x) \implies F(x + h) - F(x) = f(x + h) \cdot g(x + h) - f(x) \cdot g(x) =[/MATH]
[MATH]f(x + h) \cdot g(x + h) + \{ f(x) \cdot g(x + h) - f(x) \cdot g(x + h) \} - f(x) \cdot g(x) = [/MATH]
[MATH]\{f(x + h) \cdot g(x + h) - f(x) \cdot g(x + h)\} + \{f(x) \cdot g(x + h) - f(x) \cdot g(x)\} =[/MATH]
[MATH]g(x + h)\{f(x + h) - f(x)\} + f(x)\{g(x + h) - g(x)\} \implies[/MATH]
[MATH]\dfrac{F(x + h) - F(x)}{h} = g(x + h) \cdot \dfrac{f(x + h) - f(x)}{h} + f(x) \cdot \dfrac {g(x + h) - g(x)}{h}.[/MATH]
Now we are assuming that f(x) and g(x) are differentiable in some relevant interval (a, b), which means that, by definition, the limits of the two difference quotients on the RHS, the derivatives, exist in that interval. There is a theorem that says if f(x) and g(x) are differentiable in (a, b), then f(x) and g(x) are also continuous in that interval. Therefore, by definition, the limit of g(x + h) as h approaches zero exists and equals g(x). Moreover there is a theorem that says that if two summands each have a limit, then their sum has a limit equal to the sum of the limits. And finally there is a theorem that if two multiplicands each have a limit, then their product has a limit equal to the product of the limits.
Applying all that we get
[MATH]F’(x) = g(x)f’(x) + f(x)g’(x).[/MATH]
But who says those theorems are valid. Their proofs require other theorems. It is simply impossible to know where to start on proofs because this site is not a text book where axioms are specified and theorems worked out in a logical sequence.
JomoIn my opinion just before proving the product rule you prove the lemma showing that g(x+h) goes to g(x). Then in the proof of the product rule you state that you are using that lemma. This is how I was taught the product rule and how I taught it for decades. If I recall correctly, it is that way in all the calculus books I looked at. If it wasn't like that in the textbook I would never had said anything to Halls about it.