How do you find the derivative of #y = x^2 + x + 1 # using the limit definition?

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Answer 1 · 2016-07-11T19:50:22

I found: #y'=2x+1#

The limit definition tells us that:
#y'=lim_(h->0)(f(x+h)-f(x))/h#
where #h# is a small increment.
In our case we get:

#lim_(h->0)([(x+h)^2+(x+h)+1]-[x^2+x+1])/h=#

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

#lim_(h->0)(h(2x+h+1)/h)=lim_(h->0)(2x+h+1)=#

as #h->0# we get:

#y'=2x+1#