derivative (n) times!!!

[Math89]

hi every body

 

how can i derivative (n) times this function e^(1/x)????

 

 

thanks alot 4 all

[Dave]

Try calculating the first, second, third, fourth etc derivatives. You should be able to spot a pattern that can lead you to a general formula for the n'th derivative. From there, it's a simple matter of proving it via induction.

[Math89]

thanks dave alot

 

i try this way before post this thread

 

but i cant get the general formula

 

thx all

[EvoN1020v]

You need to use the chain rule to find the first derivative of [math]e^{\frac{1}{x}}[/math].

 

Can you tell us what you got?

[the tree]

You need to use the chain rule to find the first derivative of [math]e^{\frac{1}{x}}[/math]

You do? Isn't it just [math]\ln{x}\cdot e^\frac{1}{x}[/math]?

[uncool]

You do? Isn't it just [math]\ln{x}\cdot e^\frac{1}{x}[/math]?

 

No - ln is the integral, not the derivative, of 1/x. And in any case, you did use the chain rule to do that.

 

The problem in itself is quite difficult, because it includes three different methods of differentiation - the derivative of an exponential function, the derivative of an inverse of a power function, and the derivative of a product. The three together make this problem rather annoying at the least...

=Uncool-

[EvoN1020v]

My mistake. You don't have to use chain rule, because there's no other function in [math]\frac{1}{x}[/math].

 

Rather you have to use logarithm differiaitation.

I got the answer of [math]e^{\frac{1}{x}}(\frac{-lne}{x^2} + \frac{lne}{x})[/math].

 

It might be wrong. What do you think?

I have a question though. What is the derivative of [math]ln(e)[/math]? Because when they are together, they produce only [math]1[/math].

 

So if the derivative of ln(e) is 1, then the answer would be: [math]e^{\frac{1}{x}}(\frac{-1}{x^2} + \frac{1}{x})[/math] or rather [math]e^{\frac{1}{x}}(\frac{-x+x^2}{x^3})[/math].

[D H]

Evo, you were correct the first time: chain rule.

[math]u(x) = \frac 1 x[/math]

[math]\frac{du(x)}{dx} = -\frac 1 {x^2}[/math]

[math]f(x) = \exp\left(\frac 1 x\right) = f(u(x))[/math]

[math]\frac{df(x)}{dx} = \frac{df}{du}\,\frac{du}{dx} = -\frac 1 {x^2}\,\exp\left(\frac 1 x\right)[/math]

[EvoN1020v]

What I got at first was: [math]e^{ \frac{1}{x}} \cdot ln e \cdot \frac{-1}{x^2}[/math] according to the derivative rule: [math]a^u = a^u \cdot lna \cdot u'[/math].

 

The answer produced was: [math]-\frac{e^{\frac{1}{x}}}{x^2}[/math], which apparently is the correct answer according to the integrator tool here at http://integrals.wolfram.com/index.jsp.

 

In conclusion, the derivative of [math]e^{\frac{1}{x}}[/math] is [math]-\frac{e^{\frac{1}{x}}}{x^2}[/math].

 

 

Sorry for all the confusion, Math89.

[D H]

What I got at first was: [math]e^{ \frac{1}{x}} \cdot ln e \cdot \frac{-1}{x^2}[/math] according to the derivative rule: [math]a^u = a^u \cdot lna \cdot u'[/math].

 

But I when I checked the intergator tool here at http://integrals.wolfram.com/index.jsp, I got a completely different answer? How come?

 

You got a completely different answer because you used the wrong rule. See the corrected answer in my previous post.

[EvoN1020v]

Check it yourself DH. I just edited my post. Therefore, the derivative rule that I used was not wrong.

[EvoN1020v]

The second derivative is: [math]\frac{e^{\frac{1}{x}}(-1+2x)}{x^4}[/math].

 

Now, Math89, I hope you can detect a pattern in the derivatives?

[Math89]

thanks to every body who reply to me

 

but i know all this answer i can get 1'th derivative ,2'th derivative,3'th derivative....ect but i cant get the general formula of n'th derivative for this function

 

my greeting

[D H]

thanks to every body who reply to me

 

but i know all this answer i can get 1'th derivative ,2'th derivative,3'th derivative....ect but i cant get the general formula of n'th derivative for this function

 

From that, you should see that each of these derivatives appears to be of the form

[math]\frac {d^n}{dx^n}\exp\left(\frac 1 x\right) = p_n\left(\frac 1 x\right) \exp\left(\frac 1 x\right)[/math]

where [math]p_n(u)[/math] is a polynominal. It is obviously true for [math]n=0\text{, as\ }p_0(u)\equiv 1[/math]. Use induction to show that the general form is valid and to determine an inductive form for [math]p_n(u)[/math].

[Math89]

today i worked out this problem

 

and i get the general formula of n'th derivative for this function

 

the answer in the pic

 

thx to all

[Dave]

I'm glad it worked out okay

 

I have no idea whether that formula works or not, but if you want to make doubly sure I would try to prove it by induction.

[Math89]

thanks dave

 

but i made a prove for this problem but i haven't time to write it now