Question #8b5f0

1 Answer
sente
Oct 16, 2016

#dy/dx = cos(x)#

Explanation:

What we will need:

  • #lim_(x->0)sin(x)/x = 1#

This can be derived using the squeeze theorem together with some geometric arguments.

  • #{(lim_(x->a)f(x) = L_f), (lim_(x->a)g(x) = L_g):}=>lim_(x->a)f(x)g(x) = L_fL_g#

It can be shown using the definition of a limit that if two functions have finite limits at a point, then the limit of the sum of those functions is the sum of their limits at that point.

  • #{(lim_(x->a)f(x) = L_f), (lim_(x->a)g(x) = L_g):}=>lim_(x->a)(f(x)+g(x)) = L_f+L_g#

A similar property applies to sums.

  • #lim_(x->0)(cos(x)-1)/x = 0#

This can be derived from the first limit shown above by using some algebraic manipulation to show that #(cos(x)-1)/x = -(sin(x)/x)(sin(x)/(1+cos(x)))# and using the multiplication property shown above.

  • #d/dxf(x) = lim_(h->0)(f(x+h)-f(x))/h#

The definition of a derivative.

With the above, we have

#dy/dx = d/dxsin(x)#

#=lim_(h->0)(sin(x+h)-sin(x))/h#

#=lim_(h->0)(sin(x)cos(h)+cos(x)sin(h)-sin(x))/h#

#=lim_(h->0)(sin(x)(cos(h)-1) + cos(x)sin(h))/h#

#=lim_(h->0)(sin(x)(cos(h)-1)/h + cos(x)sin(h)/h)#

#=lim_(h->0)sin(x)(cos(h)-1)/h + lim_(h->0)cos(x)sin(h)/h#

#=sin(x)*0 + cos(x)*1#

#=cos(x)#

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