1. [SOLVED] Derivative

For every natural number n, the derivative of $x^n$ is

$(x^n)\prime=nx^{n-1}$

using the product rule: $(f g)\prime =f \prime{g}+ f {g}\prime.$
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I have worked out $u=n, v=x^{n-1}, \frac{du}{dx}=1, \frac{dv}{dx}=x^{n-2}$

$Therefore (f g)\prime =nx^{n-2}+x^{n-1}$

Is this correct so far?

2. Originally Posted by ronaldo_07
$(x^n)\prime=nx^{n-1}$

using the product rule: $(f g)\prime =f \prime{g}+ f {g}\prime.$
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I have worked out $u=n, v=x^{n-1}, \frac{du}{dx}=1, \frac{dv}{dx}=x^{n-2}$

$Therefore (f g)\prime =nx^{n-2}+x^{n-1}$

Is this correct so far?
if n is a constant ... no, it is not correct.

sure you didn't mean for $u = x^n$ instead of just $n$ ?

Do I make $x^{n-1}=(x^n)(n^-1)$

so that $u=x^n$ and $v=x^{-1}$

4. Using the product rule is useless. n is a constant and its derivative is 0 since the slope of a constant (e.g. y=1) is 0.
If you want to prove that $(x^n)\prime=nx^{n-1}$ then I fear that you have no choice but to compute
$\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{(x + \Delta x)^n - x^n}{\Delta x} = nx^{n-1}$ I guess that you can use the intuitive notion of a limit to solve this.
Expand the whole thing, simplify (substract $x^n$), divide by $\Delta x$ and compute the limit (everything goes to 0 except one term).

5. The question specifically asks to use the product rule as follows:

Use induction to prove the statement:

For every natural number n, the derivative of $x^n$ is

$(x^n)\prime=nx^{n-1}$

You will need to use the product rule: $(f g)\prime =f \prime{g}+ f {g}\prime.$

6. Originally Posted by ronaldo_07
The question specifically asks to use the product rule as follows:

Use induction to prove the statement:

For every natural number n, the derivative of $x^n$ is

$(x^n)\prime=nx^{n-1}$

You will need to use the product rule: $(f g)\prime =f \prime{g}+ f {g}\prime.$
Let P(n) be the statement we want to prove.

We proceed by induction on $n$, where $n \in \mathbb{N} = \{ 0, 1 , 2, 3, ... \}$

Clearly $P(0)$ is true, since $(x^0)' = 0 = 0x^{0-1}$ ....(of course, $x \ne 0$. if you want x to be anything it wants to be, then take the natural numbers greater than or equal to 1 in your statement)

Assume $P(n)$ is true, we show $P(n + 1)$ is true.

Now, $x^{n + 1} = x \cdot x^n = \cdots$

basically, you want to apply the product rule, and simplify to get $(n + 1)x^n$, which would be $P(n + 1)$

7. Originally Posted by Jhevon
Let P(n) be the statement we want to prove.

We proceed by induction on $n$, where $n \in \mathbb{N} = \{ 0, 1 , 2, 3, ... \}$

Clearly $P(0)$ is true, since $(x^0)' = 0 = 0x^{0-1}$ ....(of course, $x \ne 0$. if you want x to be anything it wants to be, then take the natural numbers greater than or equal to 1 in your statement)

Assume $P(n)$ is true, we show $P(n + 1)$ is true.

Now, $x^{n + 1} = x \cdot x^n = \cdots$

basically, you want to apply the product rule, and simplify to get $(n + 1)x^n$, which would be $P(n + 1)$
Basically I cant use product rule until i have applied N+1 in to the derivative right?

8. Originally Posted by ronaldo_07
Basically I cant use product rule until i have applied N+1 in to the derivative right?
i do not understand what you mean. use the product rule to differentiate $x \cdot x^n$. we can differentiate each of those terms because $P(n)$ is true. so we can validly apply the product rule here

although, i am thinking of starting with P(1) instead of P(0). what is the definition of natural numbers that you are using. including zero has some annoying cases that we may have to deal with explicitly.

9. I missread I got it thanks, We do Include 0 as a natural number in our study.

Thanks for the help

10. Originally Posted by ronaldo_07
I missread I got it thanks, We do Include 0 as a natural number in our study.

Thanks for the help
good.

in that case, you should explicitly show that x follows the rules. that is, P(1) is true.

11. Originally Posted by ronaldo_07
The question specifically asks to use the product rule as follows:

Use induction to prove the statement:

For every natural number n, the derivative of $x^n$ is

$(x^n)\prime=nx^{n-1}$

You will need to use the product rule: $(f g)\prime =f \prime{g}+ f {g}\prime.$
it never ceases to amaze me the results one can get when they post the original question in its entirety.