Question

Question #8b5f0

Answer 1

`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.

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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)`