Question

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

Answer 1

`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)`