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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,

Eus

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# Dy/dx: must dx be the independent variable?

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