How do you use the limit definition of the derivative to find the derivative of #f(x)=-2/(2x-1)#?

1 Answer
Steve M
Jan 5, 2017

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

Explanation:

The definition of the derivative of #y=f(x)# is

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

So if # f(x) = -2/(2x-1) # then;

# \ \ \ \ \ f(x+h) = -2/(2(x+h)-1) #
# :. f(x+h) = -2/(2x+2h-1) #

The numerator of the derivative will then be given by:

# f(x+h) -f(x)= -2/(2x+2h-1) - (-2/(2x-1)) #
# " "= 2 (-1/(2x+2h-1) + 1/(2x-1))#
# " "= 2 ((-(2x-1) + (2x+2h-1))/((2x-1)(2x+2h-1))) #
# " "= 2 ((-2x+1 + 2x+2h-1)/((2x-1)(2x+2h-1))) #
# " "= 2 ((2 + 2h)/((2x-1)(2x+2h-1))) #

And so the derivative of #y=f(x)# is given by:

# \ \ \ \ \ dy/dx = lim_(h rarr 0) (2 ((2 + 2h)/((2x-1)(2x+2h-1)))) / h #
# " " = lim_(h rarr 0) 2/h((2h)/((2x-1)(2x+2h-1)))#
# " " = lim_(h rarr 0) (4)/((2x-1)(2x+2h-1))#
# " " = (4)/((2x-1)(2x+0-1))#
# :. dy/dx = (4)/((2x-1)^2)#

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