# first term in the product rule

At the moment I am studying a Calculus book. This book states that if you have a function defined like this:

$f(x)=x^3$

Than if you expand that function according to the Power Rule with the Binomial Theorem the derivative would be $3x^2$. I do understand that the derivative is $3x^2$. However it does not make sense when I use that theorem and then simplify the it.

The simplification of the derivative:

$f(x)= \frac{ f(x + h)^3 - f(x)^3 }{ h }$

$f(x)= \frac{ (x + h)(x + h)(x + h) - x^{3} }{ h }$

$f(x)= \frac{ x^{2}+ 2xh + h^{2}(x + h) - x^{3} }{ h }$

$f(x)= \frac{ x^{3} + x^{2}h + 2x^{2}h + 2xh^{2} + h^{2} + h^{3}- x^{3} }{ h }$

$f(x)= 3x^{2} + 3xh + h^{2}$

So the book states, that because of that every term, except the first one, has h as a factor and therefore approaches 0. Therefore the answer is.

$f(x)= 3x^{2}$

Why is it, that only the first term is important and not the other two?

$f(x)= 3x^{2} + 3xh + h^{2}$

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You have $f(x+h)^3$ and $f(x)^3$ where you should have $f(x+h)$ (which is equal to $(x+h)^3$) and $f(x)$ (which is equal to $x^3$). Later you have $x^2+2xh+h^2(x+h)$ where you should have $(x^2+2xh+h^2)(x+h)$. And it is not at all correct to say that $f(x)$ is equal to $\dfrac{f(x+h)-f(x)}{h}$. Rather the things that are equal to each other are $f'(x)$ and $\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$. ${}\qquad{}$ – Michael Hardy Aug 24 '13 at 15:56

The derivative of $f(x)$ is denoted by $f'(x)$ and is defined as $\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}h$.

If you choose $f(x)=x^3$, we have

$f'(x)=\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}h=\lim_{h\rightarrow0} \dfrac{(x+h)^3-x^3}h=\lim_{h\rightarrow0}\dfrac{3x^2h+3xh^2+h^3}h$.

Intuitively, $\lim_{x\rightarrow a}g(x)$ is the expression to which the value of the function $g$ gets close to when evaluated at values very close to $a$ but not at $a$.

Since for all $h\ne 0$, $\dfrac{3x^2h+2xh^2+h^3}h=3x^2+3xh+h^2$, the derivative is given by $\lim_{h\rightarrow 0}3x^2+3xh+h^2$.

Now as $h$ approaches $0$, $3x^2+3xh+h^2$ gets close to $3x^2$ since $3xh+h^2$ gets close to $0$ and thus the derivative is $3x^2$.

If someone is asking you what $\lim_{h\rightarrow0}3x^2+3xh+h^2$ is, they are asking you what will $3x^2+3xh+h^2$ come close to if you substitute for $h$ values very close to $0$ (but not $0$ itself). Take an example: Find $\lim_{x\rightarrow 2} \frac{(x+2)(x-2)}{x-2}$. Now try to put values for $x$ that are very close to $2$ and see what the function, $f(x)=\frac{(x+2)(x-2)}{x-2}$ gets close to. For $x=2.1$, $f(x)=4.1$. Similarly $f(2.01)=4.01$, $f(2.001)=4.001$, $f(2.00001)=4.00001$. Also for values below $2$, $f(1.9)=3.9$, $f(1.999)=3.999$, $f(1.999999)=3.999999$. It is evident that the function $f$ is approaching $4$, as you input values close to $2$. Note that the function is itself undefined at $x=2$, but that does not matter since the limit only cares about how the function behaves near $2$ but not at $2$. So, $\lim_{x\rightarrow 2} \frac{(x+2)(x-2)}{x-2}=4$

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Oke thank you. I understand a little more. But what I did not understand is. Now as h approaches 0, $3x2+3xh+h^{2}$ gets close to $3x^{2}$ since $3xh+h^{2}$ gets close to 0 and thus the derivative is $3x^{2}$. Could you explain why this is? The character h means the infinitesimal change, I am aware, but who decided that the infinitesimal change is so little that we say it equals zero, so it cancels the other terms and therefore the first term is left. – Clifford Aug 24 '13 at 16:33
Maybe I get it. Although it is not known what the true value of h is, it is a infinitesmall change. Infinitesimall change means a change near to zero. And therefore, because it is so small we cancel the terms because multyplying by zero yields zero. But is isn't in wrong to assume that h is so small. For example what if I have a function. $f(x)=x^{2}$ And I use have these. $f(2)=2^{2} and f(3)=2^{2}$ Than h would not be so small right? Or am I missing something. – Clifford Aug 24 '13 at 16:45
@Clifford The derivative of a function $f(x)$ is formally defined as $\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}h$. There is a formal definition of the limit too, but I'm presuming you're just beginning to learn calculus and are trying to understand the ideas intuitively first. Usually, an informal introduction to calculus starts by talking about what the limit means intuitively and this is what I've tried to explain. I've edited my answer, so feel free to ask for clarification if you didn't understand something. – Alraxite Aug 24 '13 at 17:53
Thank you very much for your clear explanation. It makes sense now. – Clifford Aug 26 '13 at 8:37

First, it is not convenient to denote $$\frac{ f(x + h)^3 - f(x)^3 }{ h }$$ by $f(x)$ for two reasons: we have already defined $f$ to be the function $x\mapsto x^3$ and the variable of this function is $h$ since finally we pass to the limit $h\to0$.

Now to answer your question: if we pass to the limit $h\to 0$ the terms $3xh$ and $h^2$ vanishes and remains the term $3x^2$ which represents the derivative of $x^3$.

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the derivative of x^3 is 3x^2. – Rajath Krishna R Aug 24 '13 at 14:47
@RajathKrishnaR I fixed the typo. Thanks. – user63181 Aug 24 '13 at 15:17
Oke thank you very much. Indeed I wrote it wrong. It should be defined as a limit. – Clifford Aug 24 '13 at 16:39

It is not 'f(x)=' but its 'as limit h approaches 0'. So, after simplification you put h=0 and thus, the other terms cancel out. The book shows the method of differentiating using first principle.

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In the early development of the calculus, mathematicians were interested in the effects of "small changes" in the domain ($x$ values) of a given function on the range of the given function (the values $f(x)$ which result)

A small change was called a differential, which we will write "$\text{d}x$". In some sense this can be thought of as the value $h$ you have above, and as $\text{d}x$ gets very small, or similarly $h$ gets very small, we understand it to "go to zero." This is the reason we ignore the terms with coefficients of orders of $h$ in your equation.

Here's an interesting example from history by Euler (adapted from Dunham 2008, "The Calculus Gallery," pp.53-54), which would not meet today's standard of proof, but might give some insight into what's going on:

Euler was interested in differentials of the sine function. He started with the power series of $\sin(x)$ and $\cos(x)$, thus

$$\sin x = x-x^3/3!+x^5/5!-\cdots$$ and $$\cos x = 1-x^2/2!+x^4/4!-\cdots.$$

He then substituted $\text{d}x$ into both (which he understood to be "infinitely small" - whatever that means!):

$$\sin(\text{d}x) = \text{d}x-(\text{d}x)^3/3!+\cdots$$ and $$\cos(\text{d}x)=1-(\text{d}x)^2/2!+\cdots.$$

The higher powers of $\text{d}x$ (such as $(\text{d}x)^2$ and $(\text{d}x)^3$ etc.) were considered insignificant (since they "are zero even more so than $\text{d}x$ or a constant" - again whatever that means!). Hence, we can write (*)

$$\sin(\text{d}x)=\text{d}x$$ and $$\cos(\text{d}x)=1.$$

Euler then considered the function $y=\sin(x)$, and using the idea of the differential considered:

$$y+\text{d}y = \sin(x+\text{d}x).$$

Using a trigonometric identity this gives

$$y+\text{d}y=\sin(x)\cos(\text{d}x)+\cos(x)\sin(\text{d}x),$$ which using (*) gives

$$y+\text{d}y = \sin(x)+\cos(x)\text{d}x.$$

Subtracting $y$ from both sides then gave

$$\text{d}y = \sin(x)+\cos(x)\text{d}x-y,$$

which simplifies to $$\text{d}y = \cos(x)\text{d}x.$$

In modern notation this tells us that $$\frac{\text{d}y}{\text{d}x} = \cos(x),$$ i.e. the derivative of $y=\sin(x)$ is $\cos(x)$ !

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