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find the derivative of f(x)=x^2 + x -3 using the definition of the derivative

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\[f'(x)\]\[=\lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{[(x+h)^2 + (x+h) -3] -(x^2 + x -3)}{h}\]Can you simplify the fraction first?
I used x^2 + 2(x)(h) + h^2 + x +h -3 -x^2 -x +3 = 2x + x ??
x^2 + 2(x)(h) + h^2 + x +h -3 -x^2 -x +3 <- correct for denominator. But it's not equal to 2x + x... \[x^2 + 2(x)(h) + h^2 + x +h -3 -x^2 -x +3\]Group the like terms together: \[x^2 -x^2 + x -x -3 +3+ h^2+ 2(x)(h)+h\] Now, can you simplify the above expression

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Other answers:

basically factor out the h and H(h+2x+1) divided by the h on denominator and = h + 2x + 1?
And you got: \[\lim_{h \rightarrow 0} (h+2x+1)\]Can you evaluate the limit?
Would be 2x + 1 ??? Because h is going to 0
Indeed it is!
Holy Cow! It came together finally!! Thank you very much for your help! I would have been stuck!!!
You're welcome :)
Have a great day!!
You too! :)

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