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Dy/dx: must dx be the independent variable?

  1. Feb 26, 2008 #1


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    Hi Ho!

    If [tex]y=\sin x[/tex], [tex]\frac{dy}{dx}=\frac{d(\sin x)}{dx}=\cos x[/tex].

    If [tex]x=\sin y[/tex], [tex]y=\arcsin x[/tex], and therefore [tex]\frac{dy}{dx}=\frac{d(\arcsin x)}{dx}=\frac{1}{\sqrt{1-x^2}}[/tex].

    But, if [tex]x=\sin y[/tex], can [tex]\frac{dy}{dx}[/tex] be done as [tex]\frac{dy}{dx}=\frac{dy}{d(\sin y)}=\frac{1}{\frac{d(\sin y)}{dy}}}=\frac{1}{\cos y}=\sec y[/tex]?

    If it cannot, is it because the definition of [tex]\frac{dy}{dx}[/tex] that says that [tex]\frac{dy}{dx}=\lim_{\Delta x\rightarrow 0}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}[/tex] requires that [tex]dx[/tex] must be the independent variable and [tex]dy[/tex] must be the dependent variable? Or, is there any other reason?

    Thank you very much.

    Best regards,
  2. jcsd
  3. Feb 26, 2008 #2


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    Staff Emeritus
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    Actually, in most mathematics there is little or no distinction between "independent" and "dependent" variables. All we really need is that there be a "relation" between x and y. If y= sin(x), as long as we stay on an interval in which sine is "one-to-one", we can define the inverse function x= arcsin(y). Another way to find the derivative of x= sin(y), with respect to x, is to use "implicit differentiation": take the derivative of both sides with respect to x. 1= cos(y)dy/dx (by the chain rule) so dy/dx= 1/cos(y)= sec(y). If it bothers you that the right side is a function of y instead of x, you can change that by using trig identities, or even simpler thinking of y as an angle in a right triangle. If x= sin(y), then we can take x as the "opposite side" of a right triangle with hypotenuse 1. y is the angle opposite side x of course, so sin(y)= opposite side/near side= x/1= x. By the Pythagorean Theorem, then, the length of the "near side" is [itex]\sqrt{1- x^2}[/itex]. Since secant is defined as "hypotenuse over near side", sec(y)= [itex]1/\sqr{1- x^2}[/itex]. That is, if x= sin(y), dx/dy= [itex]1/\sqrt{1- x^2}[/itex] which is precisely the derivative formula for "arcsin(x)".
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