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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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thinker1Best ResponseYou've already chosen the best response.0
no mandw:1352457865514:dw
 one year ago

xcryptBest ResponseYou've already chosen the best response.0
that's not what I am asking though
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
do you agree that \[\huge \frac{dy}{dx} = \frac d{dx}\] they're the same. right?
 one year ago

xcryptBest ResponseYou've already chosen the best response.0
I have never the right hand side notation before
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
you're not familiar with d/dx?
 one year ago

xcryptBest ResponseYou've already chosen the best response.0
is it the same by definition?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
that's a bummer...i would assume you're at the start of your calculus class...hmm i'll try thinking of another analogy
 one year ago

xcryptBest ResponseYou've already chosen the best response.0
I'm actually doing real analysis
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
...then why are you not familiar with d/dx?
 one year ago

xcryptBest ResponseYou've already chosen the best response.0
I have never seen it before, not in my books
 one year ago

xcryptBest ResponseYou've already chosen the best response.0
But I've never taken calculus before ra so that might explain it
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
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
 one year ago

thinker1Best ResponseYou've already chosen the best response.0
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 \[d ^{n}x /d ^{n} y\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
\[\frac{dy}{dx} = \frac d{dx}\] 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
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
the derivative of a function is the rate that dependent variable changes as the independent variable varies in order
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
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....
 one year ago

xcryptBest ResponseYou've already chosen the best response.0
oh, we write f' or dy/dx instead of d/dx
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
d/dx means derivative of with respect to x so dy/dx means derivative of y with respect to x
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[y=x^2\]\[\frac{\text dy}{\text dx}=2x\]\[\frac{\text d^2y}{\text dx^2}=2\] \[\frac{\text d^2y}{\text dx}=2\text dx\]\[{\text d^2y}=2\text dx{\text dx}\]\[{\iint \text d^2y}=2\iint\text dx^2\]\[\int \text dy=2\int( x+c)\text dx\]\[y=2\frac{x^2}2+c\]\[y={x^2}+cx+d\]
 one year ago
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