anonymous
  • anonymous
This is probably a weird question but is there a reason or intuition we write higher order derivatives as \[ \frac{d^n y}{dx^n} \] instead of \[ \frac{d^n y}{d^n x} \] or something similar?
Calculus1
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
no man|dw:1352457865514:dw|
anonymous
  • anonymous
do u understand mate
anonymous
  • anonymous
that's not what I am asking though

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lgbasallote
  • lgbasallote
do you agree that \[\huge \frac{dy}{dx} = \frac d{dx}\] they're the same. right?
anonymous
  • anonymous
I have never the right hand side notation before
lgbasallote
  • lgbasallote
you're not familiar with d/dx?
anonymous
  • anonymous
no
anonymous
  • anonymous
is it the same by definition?
lgbasallote
  • lgbasallote
that's a bummer...i would assume you're at the start of your calculus class...hmm i'll try thinking of another analogy
anonymous
  • anonymous
I'm actually doing real analysis
lgbasallote
  • lgbasallote
...then why are you not familiar with d/dx?
anonymous
  • anonymous
I have never seen it before, not in my books
anonymous
  • anonymous
But I've never taken calculus before ra so that might explain it
lgbasallote
  • lgbasallote
hmm i can't think of any other way to explain the reason behind d^n y/dx^n without using d/dx...so i suppose i should explain that
anonymous
  • anonymous
actually i wanted to show u that if \[d ^{n}y /d ^{n} x\]would have meant that its reciprocal had just an inverse relationship with it but actual \[d ^{n}y /d ^{n} x\] is differently related to \[d ^{n}x /d ^{n} y\]
lgbasallote
  • lgbasallote
\[\frac{dy}{dx} = \frac d{dx}\] mainly because \[\frac{dy}{dx} \implies \frac{d}{dx} (y)\] (this is algebra) then since y is usuallyimplied in functions, it is usually ommitted..that's why they just write d/dx for dy/dx
UnkleRhaukus
  • UnkleRhaukus
the derivative of a function is the rate that dependent variable changes as the independent variable varies in order
lgbasallote
  • lgbasallote
then... let's say you want the second derivative...the second derivative would be \[\frac d{dx} (\frac{dy}{dx})\] so again...using algebra, you get \[\frac{d^2y}{dx^2}\] i don't know if that made anything clearer for you....
anonymous
  • anonymous
oh, we write f' or dy/dx instead of d/dx
lgbasallote
  • lgbasallote
d/dx means derivative of with respect to x so dy/dx means derivative of y with respect to x
anonymous
  • anonymous
ah ok
UnkleRhaukus
  • UnkleRhaukus
\[y=x^2\]\[\frac{\text dy}{\text dx}=2x\]\[\frac{\text d^2y}{\text dx^2}=2\] \[\frac{\text d^2y}{\text dx}=2\text dx\]\[{\text d^2y}=2\text dx{\text dx}\]\[{\iint \text d^2y}=2\iint\text dx^2\]\[\int \text dy=2\int( x+c)\text dx\]\[y=2\frac{x^2}2+c\]\[y={x^2}+cx+d\]

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