Understanding derivatives

Question

I don't know if this is written somewhere else. I've looked all over the internet so apologies if this has already been covered.

I'm doing Year 12 Maths in Australia for what it's worth. In our textbooks the formula to find the derivative of a function is:

If $y = ax^n$ then $\frac {\mathbb d y} {\mathbb d x} = anx^{n-1}$. I can see that this works algebraically, but..

An explanation/formula that I've found on the internet but not in my textbook is in the form: $\frac {f(x+h)-f(x)} h$.

With the latter formula, which we're not taught, I can see visually on a graph and algebraically if I make the difference between two x value an ever smaller difference like 0.000...0001 then it gives a y difference value that very closely approximates the actual tangent at a given point. Hopefully I got that correct.

What I'm hoping to get is an intuitive understanding of why this is the same (I gather not always?) as in the form $\frac {\mathbb d y} {\mathbb d x} (ax^n) = anx^{n-1}$ ?

Thanks for your help!

Answer 1

To understand such formula, it is often helpful to do an example first and then deal with the general case.

Let's take $f(x) = x^2$, an easy function. With your formula it is $f'(x) = 2x^{2-1} = 2x$

What you have found is actually $ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ .

As you can see, divding through $0$ is forbidden. We set in the function and get:

$\lim_{h \to 0} \frac{(x+h)^2-x^2}{h} = \lim_{h \to 0} \frac{x^2+2xh+h^2-x^2}{h} = \lim_{h \to 0} \, 2x+h = 2x$, which is exactly what your other formula says, too.

Actually, $ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ is the definition of the derivative. All other formula can be deduced starting from this definition.

Answer 2

It really depends on what your definition of the derivative is. There are two standard definitions given in Calculus: $$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ and $$f'(a) = \lim_{a\to x} \frac{f(x)-f(a)}{x-a}.$$

Let's suppose that $f(x) = x^n$. Then $$x^n - a^n = (x-a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-1}).$$

This gives: $$f'(a) = \lim_{x \to a} \frac{x^n - a^n}{x-a} = \lim_{x\to a}\frac{(x-a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-1})}{x-a} = $$ $$\lim_{x\to a} x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-1}$$

Now notice that since we have eliminated the denominator, we can now simply evaluate the limit of the polynomial remaining by evaluating it at $x=a$. Now each term becomes $a^{n-1}$ and there are $n$ terms. This yields:

$$f'(a) = na^{n-1}.$$

Answer 3

What your textbook gave you as a derivative is a very special case, namely the derivative of a polynomial. You can actually use the definition you found on the internet to prove the formula $\frac{\mathrm d}{\mathrm dx} x^n = nx^{n-1}$:

$$\frac{\mathrm d}{\mathrm dx} x^n = \lim_{h\to0} \frac{(x+h)^n-x^n}h \\ = \lim_{h\to0} \frac{x^n + nx^{n-1}h + \frac{n(n-1)}2 x^{n-2}h^2 + \ldots + h^n - x^n}h = \lim_{h\to0} nx^{n-1} + h\left(\frac{n(n-1)}2 x^{n-2} + \ldots + h^{n-2}\right) = nx^{n-1}$$

Answer 4

You could check it for yourself if you know the Binomial Theorem.

Just expand $(x+h)^n=x^n+nhx^{n-1}+\text {terms involving higher powers of} h$. Then cancel the $h$ from the bottom line and let $h\rightarrow0$.

Then you have your result.

Answer 5

the quotient you have is called the difference quotient. it approximates the slope of the curve/tangent line at the point.

just to be specific lets take $$f(x) = x^2$$ and look at the tangent at the point $(1,1).$ first we will look at the difference quotient $$\frac{x^2 - 1^2}{(x-1)}$$ for values of $x$ close to $1.$

$\begin{array}{|c|c|c|c|}\hline x & .9 & .99 & .999 & 1.0 & 1.001&1.01 &1.1\\\hline \frac{x^2-1}{x-1}&1.9&1.99&1.999&undefined&2.001&2.01&2.1\\ \hline\end{array}$

you can see from the table that the difference quotient approaches the value of $2$ as $x$ approaches $1$ from the left and from the right.

we say that the graph of $y = x^2$ has a slope of $2$ at the point $(1,1)$ or the derivative of $x^2$ at $x=1$ is $2.$

Answer 6

The differential quotient is $\lim_{a\to b}\frac{a^n-b^n}{a-b}$

I drop the coefficient here.

Polynomial divison:

$(a^n-b^n):(a-b)=a^{n-1}+a^{n-2}\cdot b^1+ a^{n-3}\cdot b^2+...+\cdot a\cdot b^{n-2}+b^{n-1}$

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Some steps of the polynomial division:

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There are n summands.

Inserting $b$ for $a$

${b}^{n-1}+b^{n-2}\cdot b^1+ b^{n-3}\cdot b^2+...+\cdot b\cdot b^{n-2}+b^{n-1}$

This is $n\cdot b^{n-1}=\lim_{a\to b}\frac{a^n-b^n}{a-b}$

For $a=x+h$ and $b=x$ we get

$n\cdot x^{n-1}=\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}$

Answer 7

I'm going to approach this from a less formal direction.

Suppose you have $t(x) = k * u(x)$. $t$ is a function that is another function, $u$, times a constant $k$. The slope of $t$ is then $k$ times the slope of $u$, right? Or, $t' = k * u'$.

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If you have $h(x) = f(x)*g(x)$ what happens to the slope? Well, we can figure it out formally or we can tell ourselves a little story.

Imagine if $g(x)$ did not vary -- it was equal to a constant $k$. Then the slope would be the slope of $f(x)$ times the constant $k$.

Now imagine if $f(x)$ did not vary -- it was equal to a constant $m$. Then the slope would be the slope of $g(x)$ times the constant $m$.

Now they both vary. We end up with the rule $h'(x) = k*f'(x) + m*g'(x)$ where $m=f(x)$ and $k=g(x)$, because as you move a bit to the left of $x$ both components end up moving the result up, each by a different amount.

Once we replace $m$ and $k$ with $g(x)$ and $f(x)$ respectively:

$$h'(x) = g(x)f'(x) + g'(x)f(x)$$ which is the chain rule.

Intuitively, the rule is that the slope at that point is the sum of the slopes of each component. And each component gets its slope multiplied by the current magnitude of the other component.

In a real proof, there are second-order effects that the above glosses over. A formal proof might show that the second-order effects can be neglected (that they disappear in the limit).

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Now let us apply this to $f(x) = x^2$. If $h(x) = x$, then $f(x) = h(x)*h(x)$.

We apply the chain rule: $f'(x) = h(x)h'(x) + h'(x)h(x)$.

If we know that $h'(x) = 1$ (because we know the function $y=x$ has a slope of 1), then we get:

$$f'(x) = x * 1 + 1 * x = 2x$$ We can back up to the chain rule, each of the two $x$s contribute a slope of 1, times the magnitude of the other component (which is $x$). These are then added together.

We can then look at either $x^3$ or $x^4$. For $x^4$ this is $x^2 * x^2$, so the derivative is $2x x^2 + 2x x^2 = 4x^3$. For $x^3$ this is $x * x^2$, so the derivative is $1 * x^2 + x * 2x = 3x^2$.

This pattern continues to infinity -- that the derivative of $x^n$ is $n x^{n-1}$. There are many ways to prove this, from combinatorial to calculus to induction on the chain rule.

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The derivative is often defined as a limit. But the derivative is actually a way to approximate a function with a linear one around a particular point. The $f'(7) = k$ means that close enough to $7$, the function $g(x) = k (x-7)$ is a "good" approximation of $f(x)-f(7)$.

The formalism -- as a limit -- all happened long after people where working with derivatives in the real world. The formalism put it on solid enough ground that we can work out the places where the intuitive definition is probably going to fall apart, instead of blindly walking over a logical cliff in the middle of a mathematical argument, with little ability to work out where you went wrong.

Answer 8

A lot of people have been showing you the proof using the binomial formula, but I'll give it to you in a more general way.

Let $f(x)=x^n$ for any $n\neq-1$.

Thus we know $$f(x)=e^{n\ln(x)}$$ Therefore: $$f'(x)=\frac d{dx}e^{n\ln(x)}\\f'(x)=e^{n\ln(x)}\cdot\frac d{dx}n\ln(x)\\f'(x)=x^n\cdot\frac n{x}\\f'(x)=nx^{n-1}$$ Which is a pretty neat trick, if you ask me. If you are confused, note that $$\frac d{dx}\ln(x)=\frac1{x}$$