## anonymous one year ago What's my mistake here? GIVEN f[x]=x/E^x^2 (f[ x +h] - f[x]) \ h

1. anonymous

What's my mistake here... ? I dont think this is the right answer $\frac{(x+ h)}{e^{(x+h)^{2}}} - \frac{x}{e^{x^2}} * \frac{1}{h}$ Find common denominator ... $\frac{e^{x^2}(x+ h)}{e^{x^2}e^{(x+h)^{2}}} - \frac{x e^{(x+h)^{2}} }{e^{x^2} e^{(x+h)^{2}}} * \frac{1}{h}$ $\frac{ e^{x^2} (x+ h) - x e^{(x+h)^{2}} } { e^{x^2} e^{(x+h)^{2}} } * \frac{1}{h}$ expanded $\frac{ x e^{x^2} + h e^{x^2} - x e^{x^2+2hx +h^2} } { e^{x^2} e^{x^2+2hx+h^2} } * \frac{1}{h}$ exponent laws ... $\frac{ x e^{x^2} + h e^{x^2} - x e^{x^2} e^{2hx} e^{h^2} } { e^{x^2} e^{x^2} e^{2hx} e^{h^2} } * \frac{1}{h}$ simplify and cancel out terms $\frac{ x e^{x^2} + h e^{x^2} - x e^{x^2} } { e^{x^2} e^{x^2} } * \frac{1}{h}$ $\frac{ h e^{x^2} } { e^{x^2} e^{x^2} } * \frac{1}{h}$ $\frac{ h } { e^{x^2} } * \frac{1}{h}$ $\frac{ 1 } { e^{x^2} }$

2. anonymous

Your cancelling is wrong.

3. anonymous

yeah, looking for another way to explain it.. without the derivative rules because this stupid course throws this problem at me in chapter 3 and doesnt introduce derivatives until chapter 4.. lol damn them

4. anonymous

just thought for a minute I found a less complicated way there, by getting a common denominator. my cancelation algebra is off though I guess

5. Australopithecus

Yeah that isn't the derivative of that

6. anonymous

Down the route you went. The problem is at this step: $\frac{ xe^{x^{2}}+he^{x^{2}}-xe^{x^{2}}e^{2xh}e^{h^{2}} }{ e^{x^{2}}e^{x^{2}}e^{2xh}e^{h^{2}} }*\frac{ 1 }{ h } = \frac{ x + h - xe^{2xh}e^{h^{2}} }{ e^{x^{2}}e^{2xh}e^{h^{2}}}*\frac{ 1 }{ h }$ That would be the simplification you can do, an $$e^{x^{2}}$$ cancelling out.

7. anonymous

....Not that helped any in terms of trying to do this with the limit definition, lol.

8. anonymous

lol..

9. Michele_Laino

I got this: $\large \frac{1}{h}\left( {\frac{{h{e^{{x^2}}} - x{e^{{h^2}}} - x{e^{2xh}}}}{{{e^{{x^2} + {{\left( {x + h} \right)}^2}}}}}} \right)$

10. Australopithecus

you need chain rule to solve this

11. Australopithecus

When I guided you through this the other day, Noted that the derivative of e^(f(x)) is f'(x)e^(f(x)) this is not really a derivative rule it is due to chain rule. These lectures might help. https://www.youtube.com/watch?v=8UHQhpoFF1c The solution i gave you did not require the use of any derivative rules.

12. Australopithecus

Here is also a detailed proof of chain rule http://kruel.co/math/chainrule.pdf

13. anonymous

true, I was reconstructing it, when I thought I had found a shortcut.. but I guess not.. because this route just cancels out to 0 again.

14. Australopithecus

It is because it is a composite function, it is good that you are doing this though as the ability to see functions in other functions is fundamental to being able to solve derivatives

15. Australopithecus

or rather to be able to break a function into smaller functions

16. Australopithecus

http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeProofs.aspx Pauls notes also have an explanation of chain rule that is pretty alright

17. Australopithecus

you can probably incorporate the chain rule proof to solve this problem

18. anonymous

Not that it would help, but if you still feel the need to keep trying the limit definition route, you could try this variation and see if you can come up with some fancy simplification of it: $\frac{ f(x) - f(t) }{ x-t }$ as x approaches t. Another thing to try is rewriting x as $$e^{lnx}$$ and h as $$e^{lnh }$$ and maybe try to write everything as exponents

19. Australopithecus

If you prove d/dx e^(f(x)) is f'(x)e^(f(x)) then you can say your problem is completely solved. I feel like this problem is not really something you should dwell to hard on you will likely never encounter a problem like this in real life. Although it cant hurt to learn proofs I suppose, as it will give you a deeper understanding of the mathematics.

20. anonymous

that looks like a solution actually.. prove d/dx heheh.. I have like 6 chapters of derivative related stuff coming up.. I think they just wanted me to think on this, before they give me the chain rule and power rules. but I was a bit worried if I dont get this, I wont get whats coming up. going to go watch some more vids on limits, derivatives, chain and power rules, differentials.. mind you none of those simple examples seem to deal with complex rationals and multi term exponents, so I have found myself wondering a lot, "how would I apply this?" I'm starting to see though that terms can be treated as f(x) g(x) in various ways and exponents can be broken up, so it will click soon.

21. Australopithecus

I bet you will breeze through it, calculus is actually a pretty easy topic if you have algebra mastered.