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

1 Answer
Noah G
Dec 2, 2016

#y' = 2/(2x+ 1)^2#

Explanation:

We use the formula #f'(x) = lim_(h-> 0) (f(x + h) - f(x))/h#.

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

#y' = lim_(h->0) (-1/(2(x + h) + 1) - (-1/(2x+ 1)))/h#

#y' = lim_(h->0) (-1/(2x + 2h + 1) + 1/(2x + 1))/h#

#y' = lim_(h-> 0) ((-(2x + 1) + 2 x + 2h + 1)/((2x + 2h + 1)(2x+ 1)))/h#

#y' = lim_(h->0) ((-2x - 1 + 2x + 2h + 1)/((2x + 2h + 1)(2x + 1)))/h#

#y' = lim_(h->0) (2h)/((2x + 2h + 1)(2x + 1)(h))#

#y' = lim_(h->0) 2/((2x+ 2h + 1)(2x + 1))#

We can now substitute directly.

#y' = 2/((2x + 2(0) + 1)(2x + 1))#

#y' = 2/(2x + 1)^2#

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