- #1
ThaSlave
- 1
- 0
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
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.
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.