
#1
Feb1304, 05:33 PM

P: 221

If f is a function defined by the fomula f(x)=xe^(modx), then show that f is differentiable at every point c, with
f'(c)=(mod(c) +1)e^(modx) The hint that is given is 'consider separately the cases cgreater than 0, c less than 0 and c=0 To prove that f is differemtiable at every point c, do I have to have f as 2 seorate functions mmultiplied together, and if each of them is differentiable then does it mean that f is differentiable? I tried this and it was fimne fpr the function equal to x, but I couldn't work it out for the function equal to e^(modx) . Is there a better way of proving that it is differentiable? Also, is there a reason why we have mod(c) inside the brakcets of f'(c) as opposed to just c? 



#2
Feb1304, 05:42 PM

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hint, the definition of mod x is
mod(x) = x if x +ve x if x ve when x is not zero the function is then either xe^x or xe^{x} and you should be able to state that that's clearly diffible. The problem is at x=0 you need to find lim h>0 of he^{mod(h)}/h = e^{mod(h)} as h goes to 0 from above AND below  that's the defintion of derivative. clearly the limit is 1 in either case, so you can conclude it's diffible now try and decide why the thing given is the derivative other example: mod(x) is not diffible at 0 cos the left and right limits in the limit defining the derivative are 1 and 1 resp. 



#3
Feb1304, 05:55 PM

P: 221

Okay my confusion came because I was working out the derivative of e^(modx) I used the definition of a derivative to work it out
f(c+h)  f(c) all divided by h as h tends to zero. So I'd have e^(mod(c+h))  e^(mod(c)) all divided by h I didn't know where to go from here because I couldn't split the e^(mod(c+h)) up to make the expression simpler 



#4
Feb1304, 06:13 PM

P: 221

How to tell if a function is differentiable or not
Suddenly the penny is starting to drop here thanks for your help! Gahhhh sometimes I can be so slow!




#5
Feb1304, 06:22 PM

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good, as long as you can see that anywhere away from zero, as h gets small, it must be that c and c+h are both the same sign, and you can remove the modulus entirely.




#6
Feb1404, 09:03 AM

P: 221

Hm got the first part of whast you're saying but the bit abaout why we need the c and c+h to be the same sign.......... not sure about that. Sorry, my kowledge of these things is a bit limited!




#7
Feb1404, 10:43 AM

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It helps a lot to know that both x and c+h are positive because that way e^{modc} and e^{mod(c+h)}are just e^{c} and e^{c+h}. Similarly, if c and c+h are both negative e^{modc}= e^{c} and e^{c+h}= e^{ch}. That means that you can ignore the "mod" part entirely. The important point is that if c> 0 to begin with, since we are interested in the limit as h>0 we can assume that h< c/2 so, at worst, with h= c/2 c+h= cc/2= c/2> 0. If c= 0, then c+h will be positive or negative depending on whether h is positive or negative and we have to consider both e^{modh}= e^{h} (when h is positive) and e^{modh}= e^{h} (when h is negative). 



#8
Feb1404, 06:16 PM

P: 221

Okay, I've just been looking at the questoion again and my problem is even how to get to this stage. Do I have to apply the product rule to xe^(modx) so I have to work through the full definition of the product rule? I'm using fg)'(c) = f'(c)g(c) + f(c)g'(c) , for the derivatives using the definiton of a derivative. I can't seem to get this down into the form they want. How am I supposed to be approaching this? I can see what you're saying is making a sense but it's just how to get to that bit. Do I need to use the product rule here? Just using the definiton of a derivative isn't enough, right?




#9
Feb1404, 06:24 PM

P: 221

Right, it's fine when c=0, but for the others I'm finding it hard to break it down. Right now this is hat I have, for c being greater than zero
f'(c) = e^c + e^c ((e^(h)  1) divided by h))c Can you tell me where I go from here, or if I'm on the right track? Thanks aain for your help btw. [:)] 



#10
Feb1404, 06:46 PM

P: 221

Oh hang on, just faffed about with some numbers on my calculator and is it true that e^h  1 all divided by h tends to 1 as h tends to zero? Why is this? Sorry, it's late and brain is even more off than usual!




#11
Feb1404, 07:39 PM

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because (e^h1)/h as h tends to zero is the derivative at 0....
away from zero you can just treat it as a derivative of ordinary functions. so do it for x greater than zero and for x less than zero remembering how mod(x) is defined. you will get a derivative for each case, the *best* way to write this is in the form they gave you xe^x for x positive, its derivative is xe^x+e^x by the product rule for x negative xe^{x}, the derivative is xe^x +e^{x} remember x=modx for x negative so the answer can be encapsulated as (mod)x)+1)e^mod(x) do you really need to do it from first principles? the answer should state that the limits at 0 are the same from both sides, which you seem to have mastered. 



#12
Feb1504, 11:36 AM

P: 221

Ah I get what's oing on now! I'm not sure whether we had to get it from first principles or not but I did it that way and it looks out and everything fits in fine. So thanks for your help!




#13
Feb1504, 01:19 PM

P: 221

But I thought that at c=0 we had this
lim h>0 of he^{mod(h)}/h = e^{mod(h)} as opposed to e^h  1 or was this not what you were referring to? 



#14
Feb1504, 02:42 PM

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you asked why {e^1}/h went to 1 as h went to zero, that was why i put that in there.




#15
Feb1504, 05:04 PM

P: 221

Oh it's ok, I just didn't realize why it went to 1, like if there was some prooof. But I've just looked at my notes and we seem to be able to take it as being 1 without having to go any furhter.




#16
Feb1504, 05:14 PM

P: 221

Okay, Referring to a point that was made earlier, if we have e^ mod (h + c) can we have that if c is less an zero then h tends to zero from the positive side or vice versa? Because would the mod of this then be e^(c + h). Or do we have them always both being the same sign because we'll always get the same value whether h is positive or negative here as it tends to zero?




#17
Feb1504, 05:24 PM

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I'm not sure i really understand this question.
for h sufficiently small, x+h will be negative for a negative x, so its modulus is (x+h) irrespective of whether h is positive or negative. want to check? put x = 1 and h = 1/2 and 1/2 and see it all works out. 



#18
Feb1504, 05:51 PM

P: 221

Okay, say we've got like for c less than zero
e^(c) + ce^(c)((e^)h) 1 divided by h) this is when h is negative What would happen if we had c less than zero but h positive? 


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