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#.