Let f ( x ) = 1/(2 x − 5) . According to the definition, f'(x)= lim h-->0

Question

So I have (1/(2(x+h)-5)-1/(2x-5))/h to start. I get (2x-5)2(x+h)-5 as my denominator for the upper part of the fraction. and the numerator is (2x-5)-(2(x+h)-5) from left to right. I get stuck here because I try to solve the equation and it becomes such a mess I get an answer that is wrong and can't retrace my steps. Please help!

Answer 1

Based on what you have so far, I think you are in this step

[(2x-5)-(2(x+h)-5) / (2x-5)*(2(x+h)-5) ] 1/h

Taking only the numerator we have

(2x-5)-(2(x+h)-5)

= 2x-5-(2x+2h -5)

= 2x-5-2x-2h+5

= -2h

So going back to the first expression, now we have

[-2h/ (2x-5)*(2(x+h)-5) ] 1/h

the h in -2h cancels with the h in the 1/h. So that results in

[-2/ (2x-5)*(2(x+h)-5) ]

Now we are ready to take the limit when h->0

lim h->0 [-2/ (2x-5)*(2(x+h)-5) ] = -2/ ( (2x-5)^2 )

You can verify this result using the chain rule if you already have learned it. Or using the internet :)

Answer 2

f(x) = 1/(2x - 5)

[f(x+h) - f(x)] / h = [1/(2x+2h-5) - 1/(2x-5)] / h = [(2x-5) - (2x+2h-5)] / [(h)(2x-5)(2x+2h-5)]

= -2h / [h(2x-5)(2x+2h-5)] = -2 / [(2x-5)(2x+2h-5)]

Taking the limit as h approaches zero, we get -2/(2x-5)² as f'(x).