Using limit definition of the derivative, show that (sin7x)' = 7cos7x. Thanks

You need to use derivative definition to find what derivative of
function `f(x)=sin 7x` is such that:

`f'(x) = lim_(h-gt0)
(f(x+h)-f(x))/h`

`lim_(h-gt0) (f(x+h)-f(x))/h = lim_(h-gt0) (sin
7(x+h)-sin 7x)/h`

You need to substitute 0 for h such
that:

`(sin 7x-sin 7x)/0 = 0/0`

You
should use l'Hospital's theorem to solve the llimit such
that:

`lim_(h-gt0) (sin 7(x+h)-sin 7x)/h = lim_(h-gt0) ((sin
7(x+h)-sin 7x)')/(h')`

`lim_(h-gt0) (sin 7(x+h)-sin 7x)/h =
lim_(h-gt0) 7cos7(x+h)/1`

Substituting 0 for h
yields:

`lim_(h-gt0) 7cos7(x+h)/1 = 7cos7(x+0)/1 = 7cos
7x`

Hence, evaluating the limit using definition of
derivative yields `(sin 7x)'= 7cos 7x` .