derivative of g^n(x)

[K9-47G]

This is the problem in my book. If g(x)=x/e^x, find g^(n)(x). I don't really understand what the problem is asking me to find. It is in the differentiation section of the book, if that helps at all. I think it may be asking for a formula... By the way, the n in the formula represents how many times to take the derivative of g(x).

[Dave]

What it's looking for is an explicit formula (in terms of n and x) for the n'th derivative of g.

 

How do you approach it? The way I would consider is to work out the first few terms. Also, consider writing g as [imath]g(x) = xe^{-x}[/imath]. Once you've worked out the first few terms and figured out what the pattern is, make a guess at what you think the formula is. You can then prove it using induction (which is an easy proof).

[K9-47G]

So would it be g^n(x)=g(x)+n(e^-x)..... I'm assuming that the derivative of e^-x is e^-x....

[mezarashi]

So would it be g^n(x)=g(x)+n(e^-x)..... I'm assuming that the derivative of e^-x is e^-x....

 

Don't just assume that. The derivative of exp(-x) is -exp(-x). Notice that to differentiate g(x) = xexp(-x) you will need to use the product rule. Using that, write out:

 

g'(x)

g''(x)

g'''(x)

 

and so on. Enough so that you see a pattern.

 

For example if I had the function f(x) = exp(ax)

Then

 

f'(x) = a exp(ax)

f''(x) = a^2 exp(ax)

f'''(x) = a^3 exp(ax)

 

I can conclude that f(n)(x) = a^n exp(ax). Got it?

[K9-47G]

well I did write it out, this is what I got.

g'(x)= xe^-x+e^-x

g"(x)= xe^-x+2e^-x

g'''(x)= xe^-x+3e^-x

 

so I concluded that the formula would be g^n(x)= g(x)+n(e^-x). I'm just not sure if my math is right.

[Dave]

That's not entirely correct.

 

For example:

 

[imath]g'(x) = e^{-x} - xe^{x}[/imath]

[imath]g''(x) = -2e^{-x} +xe^{x}[/imath]

 

You must make sure to observe your negative signs

 

Also, you might well see a pattern, but that certainly isn't a proof. Whilst it's an easy inductive proof, it is required to make your observation concrete.

[mezarashi]

I'm just not sure if my math is right.

 

Right off, I disagree with your g'(x). Refer to what i said about the derivative of exp(-x). Edit: or what dave just posted right now

[Dave]

so I concluded that the formula would be g^n(x)= g(x)+n(e^-x). I'm just not sure if my math is right.

 

I don't think this is a good form for the answer. You want an explicit formula in terms of x, and whilst you have one there, it's much neater to write down something like [imath]g^{(n)}(x) = nx + b[/imath] or whatever.

[K9-47G]

so e^-x derived is -e^-x.... We haven't covered that yet.

 

Now I'm getting

g'(x)=e^-x-xe^-x

g''(x)=-2e^-x+xe^-x

g'''(x)= 3e^-x-xe^-x

 

and so on, but I have no idea how to make an explicit formula out of that because the negatives are alternating.

[Dave]

Note that [imath](-1)^{n+1}[/imath] is -1 when n is even, and 1 when n is odd. It's a common thing to find.

 

Obviously if you haven't covered stuff like the chain rule and all that, you don't need to go on and prove it by induction. I had assumed that you were at some kind of 1st year university level maths or something

[K9-47G]

The inductive proof is: show true for n=1, assume true for n=k and show true for n=k+1, right?

[mezarashi]

so e^-x derived is -e^-x.... We haven't covered that yet.

 

You should have convered: d/dx ( exp(ax) ) = a exp(ax)' date=' where a is a constant. Now let a = -1.

 

and so on, but I have no idea how to make an explicit formula out of that because the negatives are alternating.

 

(-1)^n

alternates between positive and negative for increasing integer values of n.

[Dave]

That's the idea. It's blatently easy because you can just differentiate from the statement of n=k being true.

[BobbyJoeCool]

isn't [math]g^n (x)[/math] the same as saying [math](g(x))^n[/math]?

 

In which case the derivative of the function for any n can be given using the simple diferentiaition rules...

 

[math]g(x)^n=(xe^{-x})^n[/math]

 

Genera power rule states... (or the power rule with the chain rule)

 

[math]g(x)=(a)u^b, g'(x)=(ab)u^{b-1}u'[/math]

 

so we use this on the function... with..

 

[math]u=xe^{-x}[/math]

 

and u'= to the derivative of u... using the product rule...

 

[math]-xe^{-x}+e^{-x}=(1-x)e^{-x}[/math]

 

[math]g'(x)^n=n(xe^{-x})^{n-1}(1-x)e^{-x}[/math]

 

or would that be way off and too simple?

 

EDIT: all those funky f's I had were changed to g's... opps.

 

EDIT mark II: I just noticed that KG said to take the deriviatves and not Dave... so this is kinda pointless isn't it...

[K9-47G]

So would my formula be g^n(x)= -n(-1^n)e^-x+(-1^n)xe^-x. I know there must be an easier way to write that.

[K9-47G]

How do you format your math work to look bold and easier to read?

[Dave]

isn't [math]g^n (x)[/math] the same as saying [math](g(x))^n[/math']?[/math]

 

Generally, we interpret gⁿ as:

 

[math]\underbrace{g \circ \cdots \circ g}_{\text{n times}}[/math]

 

The circle indicates composition of the function. This is mainly because mathematicians are lazy, and in general we're more interested in composition than multiplication.

 

We use [imath]g^{(n)}[/imath] to denote the n'th derivative.

 

EDIT mark II: I just noticed that KG said to take the deriviatves and not Dave... so this is kinda pointless isn't it...

 

Pretty much

[BobbyJoeCool]

How do you format your math work to look bold and easier to read?

 

[math]\LaTeX[/math]

 

Link to tutorial thread

 

And I misunderstood the question... so my answer is wrong...

[Dave]

So would my formula be g^n(x)= -n(-1^n)e^-x+(-1^n)xe^-x. I know there must be an easier way to write that.

 

Yes, but you can simplify that quite a bit. Notice that by factoring the -1 out, you get the much nicer [imath]g^{(n)}(x) = (-1)^{n+1} ( ne^{-x} - xe^{-x})[/imath].

[BobbyJoeCool]

This is mainly because mathematicians are lazy' date=' and in general we're more interested in composition than multiplication.

[/quote']

 

isn't that a contradiciton of terms? lazy, but more interested in a harder way of doing things?

 

But yes, math people are lazy, who else would come up with [math]\Delta x[/math] to represent Change in x?

[Dave]

isn't that a contradiciton of terms? lazy, but more interested in a harder way of doing things?

 

Well, I suppose we just don't want to bother writing down loads of compositions, so we come up with a shorthand way of doing it. Multiplication just isn't as useful as composition

[K9-47G]

[math] g^{(n)}(x)=-n(-1^n)e^{-x}+(-1^n)xe^{-x} [/math]

[Dave]

By the way, it's not [imath]-1^n[/imath] since this is generally interpreted as [imath]-(1^n) = -1[/imath]. You'll want to write [imath](-1)^n[/imath].

[K9-47G]

ok, thanks.