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

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Answer 1 · 2017-11-08T22:37:05

Based on the limit definition of derivative:

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

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

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

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

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

#(df)/dx = 4x+1#