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

Question

Elijah Abram · Accepted Answer

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#.

Emily Ammerman · Answer

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#

Evelyn Agard · Answer

undefined To prove that (cos x)' = -sin x using the definition of derivative, we start with the definition of the derivative:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

Let f(x) = cos x. Substituting this into the definition:

(cos x)' = lim(h->0) [cos(x + h) - cos x] / h

Using the trigonometric identity for the cosine of a sum:

(cos x)' = lim(h->0) [(cos x * cos h - sin x * sin h) - cos x] / h

(cos x)' = lim(h->0) [(cos x * cos h - cos x) - (sin x * sin h)] / h

(cos x)' = lim(h->0) [(cos x * (cos h - 1) - sin x * sin h)] / h

Using the limit definition of cosine and sine functions:

(cos x)' = lim(h->0) [(cos x * (cos h - 1) - sin x * (1 - (h^2)/2) - sin x * h)] / h

(cos x)' = lim(h->0) [(cos x * (cos h - 1) - sin x + (sin x * (h^2)/2) - sin x * h)] / h

(cos x)' = lim(h->0) [(-2 * sin^2(x/2) * sin^2(h/2) - sin x * h + (sin x * (h^2)/2))] / h

As h approaches 0, sin h / h approaches 1, and (h^2)/2 approaches 0:

(cos x)' = -sin x

Therefore, (cos x)' = -sin x.