Question

Learning about derivatives, instantaneous rate of change, ..via differential quotient equation.. and limits equations, I almost have it down now, need a review of the final few posts to check my work.. thanx

Given:
f[x] = x/e^x^2


f'[x] = Lim (h->0) [(f[x+h] -f[x])/ h]

f'[x] = Lim (h->0) [ ((x+h)/(E^(x+h)^2) - (x/e^(x^2)) ) *1/h ]

Answer 1

or at the very least please send me a good link that explains how to deal with canceling variables embedded in a polynomial exponent

Answer 2

Do you have to use the definition of the derivative to solve for your given function?

derivative of e^u = d/du * e^u

for example, derivative of e^(x^3) = (3x^2)*e^(x^3)

Here's the proof for the concept if you would like:
https://www.khanacademy.org/math/differential-calculus/taking-derivatives/der_common_functions/v/proof-d-dx-e-x-e-x

Answer 3

\[f[x] = \frac{x}{e^{x^2}}\]

\[ f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right) \]

\[ f'[x] = \lim_{h \rightarrow 0} \left(
\left(
\frac{
(x+h)
}{
E^{(x+h)^2}
}
-
\frac{
x}{
E^{(x^2)}
}
\right)
* \frac{1}{h}
\right) \]

Answer 4

Unfortunately I do..

Answer 5

i have really come to admire those rules of the derivative..

Answer 6

first principles eh

Answer 7

yeah the baby steps..

Answer 8

I've been on it for 3 days.. and my OCD wont allow me to let it go

Answer 9

I cant friggin sleep, so pissed off.

Answer 10

staring at equations for days. Im missing some fundamental cancelation trick of limits.. but all the vids Im watching, I still ahvent found one that deals with a complex rational like this, and a polynomial exponent.

Answer 11

I can get there like this..
I know the answer to this is going to be
\[ f'[x] = (1-2x^2)e^{x^2}\]
and I can get there by..

g[x] = -x^2
g'[x] = -2x
h[x] = E^g[x]
h'[x] = -2xe^-x^2
f[x] = x h[x]
f'[x] = (h'[x]/g'[x]) + (f[x] * g'[x])

but I cant submit this.

Answer 12

and I cant reverse the logic, that would allow me to do that with algebra.

Answer 13

Its alright here, we can try to work through your final limit using this informal proof.

\[\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}\]

Answer 14

Although maybe amistre64 has a better way of solving this limit or problem

Answer 15

Is that the first trick of limits, that we can remove x ?

\[ f[x] = x/E^{x^2} \]
\[ f[x] = x E^{x^2} \]

Answer 16

oops .. that 's -x
\[ f[x] = x E^{-x^2} \]

Answer 17

Well this problem demonstrates product rule as well, we already solved part of the derivative, we differentiated x/e^(x^2) in regards to the x term but we still need to differentiate x/e^(x^2) in terms of e^(-x^2) that is why we are left with that limit to solve

Answer 18

This is where you left off

\[e^{-x^2}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2}- e^{-x^2}}{h}) + 1)\]

Answer 19

oh, I just found your image...

Answer 20

yeah you could potentially use that to solve your problem

Answer 21

or the limit

Answer 22

whats the strategy to get h out of that numerator ?
should that even be our goal from here?

Answer 23

Well just considering the limit:

\[\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2}- e^{-x^2}}{h}\]

Follow along with that image I posted you should be able to use it to solve this limit

Answer 24

Whats the idea behind the first part of your image where f(x) becomes f(g(x)), is that because the equation
x / e^x^2 has been split into two functions? And if so where was the split? was it at the x^2 ?

Answer 25

There is something you should note though that will likely come up the proof of d/dx e^x = e^x

by definition of e:
\[\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1\]

Answer 26

well e^(-x^2) is a composite function

Answer 27

oh is that proof because technically, we have e^0 = (almost 1)-1 and we get a something/something

Answer 28

if f(x) = e^(x) and g(x) = -x^2

then

f(g(x)) = e^(-x^2)

Answer 29

that is how the value of e is defined

Answer 30

one of the ways it is defined

Answer 31

there was another x in there though..
f[x] = x / E^(x^2)
is the x removed, or placed into f or g ?

Answer 32

oops.. thats E^-x^2

Answer 33

we pulled it out of the limit

Answer 34

because it is technically a constant

Answer 35

ah ok.. so that sits.. outside the lim definition still ? or does it now hold some value like 1?

Answer 36

or does it cancel to 0 ? I havent clicked yet on how that constant rule is actually applied.

Answer 37

just watching another vid on constant rule.

Answer 38

lol, man, nobody explains this well.. they just gloss over and say ... 'oh we can pull this out.' and they move on

Answer 39

okay, I think it goes outside ..

Answer 40

no it just sits outside the limit, if you have learned product rule this is a result of it.

By product rule:

d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)

in this problem
assuming f(x) = e^(-x^2) and g(x) = x

we solved f(x)g'(x) the limit left represents f'(x)g(x) since g(x) = x we can pull it out of the limit because it is just the original function, the limit itself represents f'(x)

Answer 41

but it hangs around..

Answer 42

the e^(x^2) just cancels out, you could cancel it out right now it wouldn't change your solution

Answer 43

so in reference to
\[\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1\]
To be clear, this means that if we encounter e^x then we have encountered the value 1? And anything attached to that or based on that can be considered to be factored with the value 1?
so E^x^2 is actually 1^2 ?

Answer 44

No it is essentially what defines what the value of e is, it comes up when you try to use the definition of a derivative to find the derivative of e^x

http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx

Answer 45

hmmm, that's an odd and curious property. :/

Answer 46

I can help you work through the informal proof if you want.

Answer 47

@amistre64 , @Hero
Might be able to help you more so with this.

Answer 48

I think if the problem was just
f[x] = e^(x^2)
then I would have no problem..
x^2 = 2x
and the derivative would be 2x * e^x^2

but I'm not sure what to do about that leading x/ ....
maybe converting it into this form and then trying to work it out...

f[x] = x e^-(x^2)

do we just put the x outside the limit from the start?

f'[x] = x Lim (h->0) [ (f[x+h] - f[x] ) /h ]

It seems like its not a problem of 2 functions, but of 3

f[x] = x h[x]
g[x] = E^h[x]
h[x] = -x^2

Answer 49

sorry thats x g[x]

Answer 50

wish you could back edit these posts

Answer 51

Not sure what you are confused about here, look what I wrote about product rule.

where g(x) = x and f(x) = e^(-x^2)
the derivative of f(x)g'(x) is just e^(-x^2)

This is shown in your problem, if you multiply e^(-x) by both terms left in the bracket you will get

your x*(limit you need to solve) + e^(-x^2)

the limit remaining is the derivative of f(x)

Answer 52

You would have this exact limit if you were trying to take the derivative with the definition using e^(-x^2)

Answer 53

If it was e^(x) and not e^(-x^2) you would have no problem because you wouldn't need chain rule

Answer 54

you are correct it is a problem of 3 functions

Answer 55

this problem requires product rule and chain rule

Answer 56

but the third function is easily differentiable so its not a problem

Answer 57

your problem is essentially

m(x)*f(g(x))

Answer 58

this is actually a good thing for you to realize since you need to be able to recognize things like this when taking derivatives

Answer 59

you need to be able to split a function into multiple functions to take a derivative in most cases

Answer 60

ah ok.. so I've been thinking g[x] = -x^2 for all cases of g[x] .. in this situation, and not seeing that g[x] is a term used to express generally a factor in an equation that varies from place to place.. where with Expression A (g[x]) * Expression B (f[x]), is in this case, x is in expression A and hence the derivative of g[x] * f[x] is in this case just E^-x^2 and in the next stage of this equation, g[x] can and will be defined to be something else, as it is applied to the general derivative equations.

Answer 61

so Ive been scratching my head, trying to interpret the language

Answer 62

I am not understanding what you are saying.

product rule is the following:
f(x)*g(x) = f'(x)g(x) + f(x)g'(x)

so if you had xln(x)

the derivative would be:

1*ln(x) + (1/x)*x

Answer 63

g(x) = x
g'(x) = 1
f(x) = ln(x)
f'(x) = 1/x

Answer 64

okay, I think I'm going to go read up more on the product rule and the chain rule again...
by the way, is this problem going to need Ln[] to finish it off at some point?

Answer 65

I posted an informal proof you could follow along with

Answer 66

to solve the limit

Answer 67

I was just saying before, that I wasn't understanding what and how g(x) was getting defined. And I just realized it was being defined on the fly, and changing in context of the problem as it evolved.

Answer 68

so as you might understand from that last statement.. the informal proof had this mysterious g(x) throughout it.. and it wasn't intuitively clicking how that was defined and applied.

Answer 69

\[\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h} = \lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h}*\frac{-(x+h)^2 - x^2}{-(x+h)^2 - x^2}\]


do you understand that

m(x) = e^(-x^2) is a composite function g(x) = -x^2, f(x) = e^x
therefore,

m(x) = f(g(x)) = e^(g(x)) = e^(-x^2)

Answer 70

I think you've multiplied by 1, using the exponents to construct the 1.. you've chosen -x^2, so it looks like some kind of conjugate, constructed from the exponents.

Answer 71

yes, I understand on the m(x) .. thnx

Answer 72

\[=(\lim_{h \rightarrow 0} \frac{e^{-(-x+h)^2}-e^{-x^2}}{-(x+h)^2 - x^2})*(\lim_{h \rightarrow 0} \frac{-(x+h)^2 - x^2}{h})\]

Answer 73

oops

Answer 74

I made a mistake

Answer 75

that should be +x^2

Answer 76

because - (-x^2) is + x^2

Answer 77

you should be able to solve the limit with that

Answer 78

okay.. I'll have a go from there..
I got some visitors that just came in.. so give me an hour or so

Answer 79

ok cool, they left.. I told em I was busy

Answer 80

Are you saying this equation should look like this?

\[
\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h}*\frac{-(x+h)^2 + x^2}{-(x+h)^2 + x^2} \]

Answer 81

Here

\[(\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{-(x+h)^2+ x^2})(\lim_{h \rightarrow 0} \frac{-(x+h)^2+ x^2}{h})\]

Answer 82

ah because of the - in the top term .. cool

Answer 83

yeah then it splits into two limits

Answer 84

just solve them now they should be solvable

Answer 85

you will need to expand

Answer 86

everything

Answer 87

that will be amazing

Answer 88

Im excited

Answer 89

\[ (\lim_{h \rightarrow 0}
\frac{
e^{-(x+h)^2}-e^{-x^2}
}{
-(x+h)^2+ x^2
}
)
(\lim_{h \rightarrow 0}
\frac{
-(x+h)^2+ x^2
}{
h
}
) \]

\[ (\lim_{h \rightarrow 0}
\frac{
e^{-x^2-2hx-h^2}-e^{-x^2}
}{
-x^2-2hx-h^2+ x^2
}
)
(\lim_{h \rightarrow 0}
\frac{
-x^2-2hx-h^2+ x^2
}{
h
}
) \]

so far ...

Answer 90

cancel stuff out

Answer 91

\[(\lim_{h \rightarrow 0}
\frac{
e^{-x^2-2hx-h^2}-e^{-x^2}
}{
-2hx-h^2
}
)
(\lim_{h \rightarrow 0}
\frac{
-2hx-h^2
}{
h
}
)\]

... I think I'm going the wrong way with this.. should it be canceling within the exponent of the numerator somehow?

Answer 92

this looks like it will zero out again.

maybe if I multiply by log[] / log[] ?

I dont know how else to get access to those exponents.. I can split them if they are

E^(a + b)

but

E^(a - b)

is a little trickier.

Answer 93

I'll read on L hopitals.
and the chain rule..

I cant help wondering if is requires some kind of Log[] / Log [] to pull that exponent out down out of the stratosphere. so we can get access to the terms.

Answer 94

those negative exponents are hard to work with.. I know we can use a square root rule, for square roots, and Im not sure about higher degrees, but does that same idea apply to a polynomial in the exponent?

Answer 95

maybe we can treat them like exponential functions in the global scope, and just throw away the low degree terms.

Answer 96

or maybe we run the limit just on the exponent, and then after knowing which way the exponent goes.. we can then say what it will do as x when in e^(x)

Answer 97

http://www.wolframalpha.com/input/?i=lim+%28h-%3E0%29+%28+e^%28-%28x%2Bh%29^2%29+-+e^%28-x^2%29+%29%2F%28+-%28x%2Bh%29^2+%2B+x^2%29