Using the definition of derivative, how do you prove that (cos x)' = -sin x?

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Answer 1 · 2015-04-20T13:38:17

Remembering the cosine difference-to-product formula, that says:

$cosalpha-cosbeta=-2sin((alpha+beta)/2)sin((alpha-beta)/2)$

and the fundamental limit:

$lim_(xrarr0)sinx/x=1$ , or in the most general writing:

$lim_(f(x)rarr0)sinf(x)/f(x)=1$

than:

$f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h=lim_(hrarr0)(cos(x+h)-cosx)/h=$

$=lim_(hrarr0)(-2sin((x+h+x)/2)sin((x+h-x)/2))/h=$

$=lim_(hrarr0)(-2sin(x+h/2)sin(h/2))/h=$

$=lim_(hrarr0)(-sin(x+h/2)sin(h/2))/(h/2)=$

$=lim_(hrarr0)-sin(x+h/2)*lim_(hrarr0)sin(h/2)/(h/2)=-sinx*1=-sinx$ .

Answer 2 · 2015-04-20T14:57:33

use the formula for the cosine of a sum:
$cos(a+b)= cosacosb-sinasinb$

together with the fundamental tigonometric limits:

$lim_(hrarr0)sinx/x = 1$ and

$lim_(hrarr0)(cosx-1)/x = 0$

For $f(x)=cosx$ , we get:

$f'(x)= lim_(hrarr0)(cos(x+h)-cosx)/h$

$color(white)"sssss"$ $= lim_(hrarr0)(cosxcosh-sinxsinh-cosx)/h$

$color(white)"sssss"$ $= lim_(hrarr0)(cosxcosh-cosx-sinxsinh)/h$

$color(white)"sssss"$ $= lim_(hrarr0)(cosx(cosh-1)/h-sinxsinh/h)$

$color(white)"sssss"$ $= lim_(hrarr0)cosxlim_(hrarr0)(cosh-1)/h-lim_(hrarr0)sinxlim_(hrarr0)sinh/h)$

$color(white)"sssss"$ $= (cosx)(0)-(sinx)(1) = -sinx$