How do you find the derivative of #4x^2 -1# using the limit definition?

1 Answer
Steve M
Nov 15, 2016

# d/dx(4x^2-1)=8x#

Explanation:

By definition of the derivative # f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #
So Let # f(x) = 4x^2 - 1 # then;

# f'(x)=lim_(h rarr 0) ( {4(x+h)^2-1 } - {4x^2-2 } ) / h #
# f'(x)=lim_(h rarr 0) ( {4(x^2+2xh+1)-1 } - {4x^2-2 } ) / h #
# f'(x)=lim_(h rarr 0) ( 4x^2+8xh+4-1 - 4x^2 -2 ) / h #
# f'(x)=lim_(h rarr 0) ( 8xh+1 ) / h #
# f'(x)=lim_(h rarr 0) ( 8x+1/h ) #
# f'(x)=lim_(h rarr 0) ( 8x ) +lim_(h rarr 0)(1/h ) #
# f'(x)=8x + 0#
# f'(x)=8x#

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