Calculus operators

Please help me to clear why dy/dx is considered to be as a ratio tho d/dx is itself a operator to find the variation with respect to x .. can it be extended upto higher derivatives and can they still be used as mere ratios for eg d^2y/(dx)^2 say some constant c then can we use (dx)^2/d^2y as 1/c .. !! i know that the last doesnt makes any sense but if for first i.e. dy/dx is say c then dx/dy is 1/c ..... then what follows after that which forces the higher derivatives not to behave the same as for first derivatives..!! any help is appreciated.. !! :)

#Calculus #Opinions

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Comments

As Far As i know !! basically derivatives are not ratios !!! they are operators Or function performer !!! so u can not say them a ratio !! means + ( addition ) is an operator w.r.t. addition !!! so as the meaning of dx/dy woul change drastically !!! because it would change the dependent variable w.r.t independent variable !! so one can nevr say dy/dx =1/dx/dy simple it means !! and also it describes its deflecting behaviour for highr derivatives

basit azhar - 7 years, 6 months ago

had done many problems with single derivatives where dy/dx is considered as a ratio.. for finding dx/dy .. we simply reciprocate the dy/dx value.. i m asking y it's considered as a ratio.. ?

Ramesh Goenka - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$