How do you use the limit definition of the derivative to find the derivative of #f(x)=-3x^2+4#?

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Answer 1 · 2016-10-05T10:39:21

#f'(x)=-6x#

The derivative of a function #f(x)# is definited as:

Given #f(x): RR->RR#

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

Therefore, given:

#f(x)=-3x^2+4#

#f'(x)=lim_(h->0)((-3(x+h)^2+4)-(-3x^2+4))/h=#

#=lim_(h->0)(-3(x^2+2hx+h^2)+4+3x^2-4)/h=#

#=lim_(h->0)(cancel(-3x^2)-6hx-3h^2cancel(+4)cancel(+3x^2)cancel(-4))/h=#

#=lim_(h->0)-3cancel(h)(2x+h)/cancel(h)=-6x#