A lot of web pages/books show how to use cosx=sin(Pi/2-x) and the chain rule to prove that the derivative of(adsbygoogle = window.adsbygoogle || []).push({});

cosx=-sinx. My question is how to use this identity and the defintion of the derivative to prove the same thing.

Or whether it is at all possible. Seeing that i get dy/dx=(cos(x+h)-cosx)/h = (sin(Pi/2-x-h)-sin(Pi/2-x))/h =

(sin(Pi/2-x)cosh-cos(pi/2-x)sinh-sin(pi/2-x))/h. The first term here is the problem, namely sin(Pi/2-x)cosh/h, since this needs to be equal to sin(Pi/2-x) for this to work. Is there another way around?

Another way of thinking is to say that cosh goes to 1 as h goes to 0, but using the same reasoning with sinh leaves me with dy/dx=0.

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# Def. of derivative and cosx=sin(Pi/2-x) to prove y'=-sinx

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