Using the limit definition, how do you find the derivative of #f(x)=4x^2 -1#?

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Answer 1 · 2016-01-29T12:29:35

We get #f`(x)=8x# using our definition:

Let us start with our definition:
#f`(x)=lim_(h->0)(f(x+h)-f(x))/h#

where #h# is a small increment.
We can use our function into this definition:
#f`(x)=lim_(h->0)((4(x+h)^2-1)-(4x^2-1))/h=#

#=lim_(h->0)((4x^2+8xh+4h^2-1-4x^2+1)/h)=#

#=lim_(h->0)((8xh+4h^2)/h)=#

#=lim_(h->0)(4h(2x+h)/h)=#
cancel #h#
#=lim_(h->0)(4(2x+h))=#
as #h->0# we have:
#f`(x)=lim_(h->0)(4(2x+h))=8x#