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

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### 2 Answers

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` .**

By definition, f'(x) = lt h [f(x+h)-f(x)}/h , h->0

=>(Sin7x)' = lt{Sin7(x+h)-sin 7x}/h

=> (sin7x)' = lt{sin 7xcos7h+cos7xsin7h - sin 7x}/h

=> (sin7x)' = lt h Sin7x ( cos7h - 1)/h + ltcos7x (sin 7h)/h

=> sin(7x)' = 0 + cos7x * 7, as Lt cos7h -1)/h = 0 and lt sin 7h/h = lt (sin7h)/7h =1.

**=>(sin 7x)' = 7cos7x.**