How do you find the derivative using limits of #f(x)=2x^2+x-1#?

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Answer 1 · 2017-12-03T04:09:44

4x+1

#lim_(h->0)(f(x+h)-f(x))/h#

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

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

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

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

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

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

#lim_(h->0)(h(4x+2h+1))/h#

#lim_(h->0)(4x+2h+1)#

#lim_(h->0)(4x+2h+1)=4x+1#