How do you find the derivative of 3x^2-5x+23x2−5x+2 using the limit definition?

1 Answer
Steve M
Dec 3, 2016

f'(x)=6x-5

Explanation:

By definition of the derivative f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h
So with f(x) = 3x^2 - 5x + 2 we have;

f'(x)=lim_(h rarr 0) ( {3(x+h)^2-5(x+h)+2 } - {3x^2 - 5x + 2 } ) / h
:. f'(x)=lim_(h rarr 0) ( {3(x^2+2hx+h^2)-5(x+h)+2 } - {3x^2 - 5x + 2 } ) / h
:. f'(x)=lim_(h rarr 0) ( {3x^2+6hx+3h^2-5x-5h+2 - 3x^2 + 5x - 2 } ) / h

:. f'(x)=lim_(h rarr 0) ( {6hx+3h^2-5h } ) / h

:. f'(x)=lim_(h rarr 0) ( 6x+3h-5 )

:. f'(x)=6x-5
:. f'(x)=6x-5

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