## anonymous one year ago 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-&gt;0) [(f[x+h] -f[x])/ h] f'[x] = Lim (h-&gt;0) [ ((x+h)/(E^(x+h)^2) - (x/e^(x^2)) ) *1/h ]

1. anonymous

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

2. jigglypuff314

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

3. anonymous

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

4. anonymous

Unfortunately I do..

5. anonymous

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

6. amistre64

first principles eh

7. anonymous

yeah the baby steps..

8. anonymous

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

9. anonymous

I cant friggin sleep, so pissed off.

10. anonymous

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.

11. anonymous

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.

12. anonymous

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

13. Australopithecus

Its alright here, we can try to work through your final limit using this informal proof. $\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}$

14. Australopithecus

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

15. anonymous

Is that the first trick of limits, that we can remove x ? $f[x] = x/E^{x^2}$ $f[x] = x E^{x^2}$

16. anonymous

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

17. Australopithecus

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

18. Australopithecus

This is where you left off $e^{-x^2}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2}- e^{-x^2}}{h}) + 1)$

19. anonymous

oh, I just found your image...

20. Australopithecus

yeah you could potentially use that to solve your problem

21. Australopithecus

or the limit

22. anonymous

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

23. Australopithecus

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

24. anonymous

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 ?

25. Australopithecus

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: $\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1$

26. Australopithecus

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

27. anonymous

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

28. Australopithecus

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

29. Australopithecus

that is how the value of e is defined

30. Australopithecus

one of the ways it is defined

31. anonymous

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

32. anonymous

oops.. thats E^-x^2

33. Australopithecus

we pulled it out of the limit

34. Australopithecus

because it is technically a constant

35. anonymous

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

36. anonymous

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

37. anonymous

just watching another vid on constant rule.

38. anonymous

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

39. anonymous

okay, I think it goes outside ..

40. Australopithecus

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)

41. anonymous

but it hangs around..

42. Australopithecus

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

43. anonymous

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 ?

44. Australopithecus

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

45. anonymous

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

46. Australopithecus

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

47. Australopithecus

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

48. anonymous

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

49. anonymous

sorry thats x g[x]

50. anonymous

wish you could back edit these posts

51. Australopithecus

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)

52. Australopithecus

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

53. Australopithecus

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

54. Australopithecus

you are correct it is a problem of 3 functions

55. Australopithecus

this problem requires product rule and chain rule

56. Australopithecus

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

57. Australopithecus

58. Australopithecus

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

59. Australopithecus

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

60. anonymous

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.

61. anonymous

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

62. Australopithecus

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

63. Australopithecus

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

64. anonymous

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?

65. Australopithecus

I posted an informal proof you could follow along with

66. Australopithecus

to solve the limit

67. anonymous

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.

68. anonymous

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.

69. Australopithecus

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

70. anonymous

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.

71. anonymous

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

72. Australopithecus

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

73. Australopithecus

oops

74. Australopithecus

75. Australopithecus

that should be +x^2

76. Australopithecus

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

77. Australopithecus

you should be able to solve the limit with that

78. anonymous

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

79. anonymous

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

80. anonymous

Are you saying this equation should look like this? $\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h}*\frac{-(x+h)^2 + x^2}{-(x+h)^2 + x^2}$

81. Australopithecus

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

82. anonymous

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

83. Australopithecus

yeah then it splits into two limits

84. Australopithecus

just solve them now they should be solvable

85. Australopithecus

you will need to expand

86. Australopithecus

everything

87. anonymous

that will be amazing

88. anonymous

Im excited

89. anonymous

$(\lim_{h \rightarrow 0} \frac{ e^{-(x+h)^2}-e^{-x^2} }{ -(x+h)^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ -(x+h)^2+ x^2 }{ h } )$ $(\lim_{h \rightarrow 0} \frac{ e^{-x^2-2hx-h^2}-e^{-x^2} }{ -x^2-2hx-h^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ -x^2-2hx-h^2+ x^2 }{ h } )$ so far ...

90. Australopithecus

cancel stuff out

91. anonymous

$(\lim_{h \rightarrow 0} \frac{ e^{-x^2-2hx-h^2}-e^{-x^2} }{ -2hx-h^2 } ) (\lim_{h \rightarrow 0} \frac{ -2hx-h^2 }{ h } )$ ... I think I'm going the wrong way with this.. should it be canceling within the exponent of the numerator somehow?

92. anonymous

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.

93. anonymous

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.

94. anonymous

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?

95. anonymous

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

96. anonymous

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)

97. Australopithecus
98. anonymous

can you give me a screen shot, wolfram alpha crashes my computer if I go anywhere near it.. I think the people who run my course put some widget into chrome that kills it.

99. anonymous

if not that's okay.

100. Australopithecus

101. Australopithecus

oh crud I just realized

102. Australopithecus

we can solve this

103. Australopithecus

really easily using the definition of e

104. anonymous

damn, how the hell does it get that?

105. Australopithecus

pull e^(-x^2) out of the limit what do you get?

106. Australopithecus

I spaced this is easily solvable using the definition of e

107. anonymous

do we pull e^(-x^2) from the top and bottom? or just the top?

108. Australopithecus

what?

109. Australopithecus

just factor out e^(-x^2)

110. Australopithecus

and you end up with the definition of e which equals 1 because -2xh - h^2 is essentially the same as just h, as h approaches 0

111. anonymous

zero out h?

112. Australopithecus

No I am saying, $\lim_{h \rightarrow 0} \frac{e^{-2x - h^2} -1}{-2xh - h^2} = \lim_{h \rightarrow 0} \frac{e^{h(-2x - h)} -1}{h(-2x - h)} =\lim_{h \rightarrow 0} \frac{e^{h} -1}{h} = 1$

113. Australopithecus

114. Australopithecus

just factor out e^(-x^2) and the rest turns to 1

115. Australopithecus

plug it back into the original equation you had in place of the limit and it is solved

116. Australopithecus

You have to accept the assumption that $h \approx h(2x - h)$

117. Australopithecus

if you multiply something by a really small number you are going to get a really small number

118. anonymous

$\lim_{h \rightarrow 0} \frac {e^{-(x + h)^2} - e^{-x^2}} {-(x+h)^2 + x^2}$ $\lim_{h \rightarrow 0} \frac {e^{-(x^2 + 2hx +h^2)} - e^{-x^2}} {-(x^2+2hx +h^2) + x^2}$ $\lim_{h \rightarrow 0} \frac {e^{-x^2 - 2hx -h^2} - e^{-x^2}} {-x^2-2hx -h^2 + x^2}$ $\lim_{h \rightarrow 0} \frac {e^{-x^2}*(1/e^{2hx})*(1/e^{h^2}) - e^{-x^2}} {-x^2-2hx -h^2 + x^2}$ $e^{-x^2} \lim_{h \rightarrow 0} \frac { (1/e^{2hx})*(1/e^{h^2}) - e^{-x^2}} {-2hx -h^2 }$

119. Australopithecus

the rest is equal to 1 by the definition of the value of e

120. anonymous

so e^2hx is .. e^2*0*x = e^0 = 1 ? you mean?

121. anonymous

but dont we still have the problem of -2hx - h^2 = -2*0*x - 0*0 = 0 ?

122. anonymous

in the denominator

123. Australopithecus

no, I am simply saying $\lim_{h \rightarrow 0}\frac{e^{-2hx - h^2}-1}{-2hx - h^2} = \lim_{h \rightarrow 0}\frac{e^{h}-1}{h} = 1$

124. Australopithecus

I can't make it any more clear than this

125. Australopithecus

$-2xe^{-x^2}$ is what you get after canceling out h and solving the limit on the left

126. anonymous

sorry, you need to be that clear with me.. :)

127. Australopithecus

128. anonymous

yeah ... actually yeah.. I can feel the air in my lungs again .

129. Australopithecus

ok so the problem is solved are you content?

130. Australopithecus

you can do the final bit of algebra now

131. anonymous

-2hx-h^2 is equivalent to h in this situation

132. Australopithecus

I think this is what the other guy was trying to point out to us earlier

133. Australopithecus

yes

134. Australopithecus

in the original question but I hadn't really made that connection as of yet

135. anonymous

yeah.. ah ah..! okay.. thats' what that pauls website was trying to say too I think.. they mentioned this.. but it went over my head

136. Australopithecus

so you can solve this now?

137. anonymous

I think so.. , so I just have to reduce to I find that ahh.. property/ equation, Im not sure what you call that class of equations.. signatures or something? and then from there.. we have it.. just pull all the terms back in that were placed outside of the limit.. and you have your equation.

138. Australopithecus

Just try to write out the solution

139. anonymous

$e^{-x^2} \lim_{h \rightarrow 0} \frac { e^{-2hx-h^2} - e^{-x^2}} {-2hx -h^2 } = \lim_{h \rightarrow 0} \frac { e^{h} -1} {h } = 1$ amI on the right track here?

140. Australopithecus

not really

141. Australopithecus

the limit becomes 1 not the entire expression

142. anonymous

$f[x] = \frac{x}{e^{x^2}}$ $f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right)$ $f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ e^{(x+h)^2} } - \frac{ x}{ e^{x^2} } \right) * \frac{1}{h} \right)$ $f'[x] = \lim_{h \rightarrow 0} \left( \left( (x+h) e^{-(x+h)^2} - x e^{-x^2} \right) * \frac{1}{h} \right)$ $f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{-x^2-2hx-h^2} + h e^{-x^2-2hx-h^2} - x e^{-x^2} \right) * \frac{1}{h} \right)$ $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{-2hx-h^2} + h e^{-2hx-h^2} - x \right) * \frac{1}{h} \right) \right)$ $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} }{h} + \frac{ h e^{-2hx-h^2} }{h} - \frac{ x }{h} \right) \right)$ $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} + \frac{ e^{-2hx-h^2} }{1} \right) \right)$ $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{-2hx-h^2} }{1} \right)$ Limit of 2nd term = 1 $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + 1 \right)$ Factor out x $f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{-2hx-h^2} - 1 }{h} \right) + 1 \right)$ Very close to the definition for e $\lim_{h \rightarrow 0} \frac{e^h -1}{ h} = 1$ $f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h} * \frac{-2x-h}{-2x-h} \right) + 1 \right)$ $f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h(-2x-h)} * \lim_{h \rightarrow 0} ({-2x-h}) \right) + 1 \right)$ So by definition of E and the constant law $f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h} - 1 }{h} * \lim_{h \rightarrow 0} ({-2x-h}) \right) + 1 \right)$ So by definition of E and the constant law $f'[x] = x e^{-x^2} \left( 1 * ({-2x-0}) + 1 \right)$ $f'[x] = x e^{-x^2} \left( -2x+1 \right)$ $f'[x] = e^{-x^2} \left( -2x^2+x \right)$ $f'[x] = x e^{-x^2}-2x^2e^{-x^2}$

143. anonymous

SOOO close.. but I factored an extra x in there somehow.. it must be something to do with how I pulled the variables out.. I guess you need to make sure that there are brackets around the different levels. And hence why they're such sticklers for making sure the lim h->0 is carried through on all the stages.

144. anonymous

btw.. thank you, thank you, to everyone who helped me get this far..

145. anonymous

GIVEN $f[x] = \frac{x}{e^{x^2}}$ FIND $f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right)$ Substitute function $f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ e^{(x+h)^2} } - \frac{ x}{ e^{x^2} } \right) * \frac{1}{h} \right)$ Move into numerator $f'[x] = \lim_{h \rightarrow 0} \left( \left( (x+h) e^{-(x+h)^2} - x e^{-x^2} \right) * \frac{1}{h} \right)$ Expand so we can isolate the first term e^-x^2 $f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{-x^2-2hx-h^2} + h e^{-x^2-2hx-h^2} - x e^{-x^2} \right) * \frac{1}{h} \right)$ Pull e^(-x^2) out of the limit. $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{-2hx-h^2} + h e^{-2hx-h^2} - x \right) * \frac{1}{h} \right) \right)$ Distribute 1/h $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} }{h} + \frac{ h e^{-2hx-h^2} }{h} - \frac{ x }{h} \right) \right)$ Simplify and combine like terms $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} + \frac{ e^{-2hx-h^2} }{1} \right) \right)$ Isolate the 2nd term into it’s own limit and remove from the main limit. $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{-2hx-h^2} }{1} \right)$ Simplify 1/e^(h=0) = 1 $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + 1 \right)$ Factor out x $f'[x] = e^{-x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{-2hx-h^2} - 1 }{h} \right) \right) + 1 \right)$ Very close to definition for e $\lim_{h \rightarrow 0} \frac{e^h -1}{ h} = 1$ neutralize the limit, by making h term = h term in numerator and denominator so we have equivalence to E definition. $f'[x] = x e^{-x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h} * \frac{-2x-h}{-2x-h} \right) \right) + 1 \right)$ Simplify and isolate the new term we used to create a 1. $f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h(-2x-h)} - 1 }{h(-2x-h)} * \lim_{h \rightarrow 0} ({-2x-h}) \right) \right) + 1 \right)$ So by definition of E and the constant law we have two final limits. $f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h} - 1 }{h} * \lim_{h \rightarrow 0} ({-2x-h}) \right) \right) + 1 \right)$ Now we just simplify our expression to the derivative. $f'[x] = e^{-x^2} \left( \left( x \left( 1 * (-2x-0) \right) \right) + 1 \right)$ $f'[x] = e^{-x^2} \left( \left( x \left( -2x \right) \right) + 1 \right)$ $f'[x] = e^{-x^2} \left( -2x^2 + 1 \right)$ OMG!! I dont believe it! Has it been 4 days and 40+ hours!! THATS IT!! $f'[x] = e^{-x^2} \left( 1-2x^2 \right)$ $f'[x] = e^{-x^2}-2x^2e^{-x^2}$

146. hartnn

excellent work hughfuve ^^ ! :)