How do you use the limit definition to compute the derivative, f'(x), for $f(x)=cos(3x)$ ?

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Answer 1 · 2015-02-20T10:22:17

The definition of derivative in a generic point $(x,f(x))$ is:

$lim_(hrarr0)(f(x+h)-f(x))/h=f'(x)$ .

So:

$lim_(hrarr0)(cos(3(x+h))-cos3x)/h=lim_(hrarr0)(cos(3x+3h)-cos3x)/h=$

than, using the formula sum-to-product:

$costheta-cosphi=-2sin((theta+phi)/2)sin((theta-phi)/2)$ ,

$=lim_(hrarr0)(-2sin((3x+3h+3x)/2)sin((3x+3h-3x)/2))/h=$

$=lim_(hrarr0)(-2sin((6x+3h)/2)sin(3/2h))/h=$

$lim_(hrarr0)-2sin(3x+3/2h)*(sin(3/2h)/(3/2h))*3/2=$

The limit of second parenthesis is a similar to the fundamental limit:

$lim_(xrarr0)sinx/x=1$ That can becomes:

$lim_(f(x)rarr0)sinf(x)/f(x)=1$ ,

so:

$=-2sin3x*3/2=-3sin3x$ .

How do you use the limit definition to compute the derivative, f'(x), for f(x)=cos(3x)f(x)=cos(3x)?