I am a bit confused on one part of the derivation of the catenary equation. At one point my book says ds(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}= dx^{2}+ dy^{2}and thus [itex] \frac{ds}{dx}=\sqrt{1 + {y'}^2} [/itex].

however that doesn't seem very rigorous to me and i am a little wary of accepting that explanation. i know that [itex] s = \sqrt{{x'}^2 + {y'}^2} [/itex] so i tried to take the derivative of this with respect to x in order to hopefully obtain the same thing as above. so from the chain rule i have [itex] \frac{1}{2\sqrt{{x'}^2 + {y'}^2}} [/itex] and now i have to take the derivative of (x'^{2}+ y'^{2}) with respect to x and here is where i am having trouble.

i have [itex] \frac{d}{dx}{x'}^2 = 2x' \frac{d}{dx}\frac{dx}{dt} = 2x' \frac{d}{dt}\frac{dx}{dx} = 0 [/itex]. then i have [itex] \frac{d}{dx}{y'}^2 = 2y' \frac{d}{dt}\frac{dy}{dx} = 2y'y''x' [/itex] which does not seem right to me. it seems like this way should work but i am just confusing the chain rule in this last portion. can someone help straighten this out? thanks.

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# Derivation of equation for catenary

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