How do you find the derivative of #(x+1)/(x-1)# using the limit definition?

Question

Aurora Alvarado · Accepted Answer

#f'(x) = -2/(x-1)^2# #f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h# #=lim_(hrarr0)((x+h+1)/(x+h-1) - (x+1)/(x-1))/h# #=lim_(hrarr0) 1/h(((x+h+1)(x-1) - (x+h-1)(x+1))/((x+h-1)(x-1)))# #=lim_(hrarr0) 1/h((x^2+xh-h-1 - x^2 -xh - h + 1)/((x+h-1)(x-1)))# #=lim_(hrarr0) 1/h((-2h)/((x+h-1)(x-1)))# #=lim_(hrarr0)(-2)/((x+h-1)(x-1)) = (-2)/((x-1)(x-1)) = -2/(x-1)^2#

Adam Adkins · Answer

undefined To find the derivative of (x+1)/(x-1) using the limit definition, we use the definition of the derivative:

f'(x) = lim(h->0) [f(x + h) - f(x)] / h

Substitute f(x) = (x+1)/(x-1) into the definition:

f'(x) = lim(h->0) [((x+h)+1)/((x+h)-1) - (x+1)/(x-1)] / h

Simplify the expression:

f'(x) = lim(h->0) [(x+h+1)(x-1) - (x+1)(x+h-1)] / ((x+h-1)(x-1)(h))

Expand and simplify the numerator:

f'(x) = lim(h->0) [(x^2 - x + hx + h - x + 1 - x^2 - hx - x - h + 1)] / ((x+h-1)(x-1)(h))

f'(x) = lim(h->0) [2h] / ((x+h-1)(x-1)(h))

Cancel out common terms:

f'(x) = lim(h->0) [2] / ((x+h-1)(x-1))

Now, substitute h = 0:

f'(x) = 2 / ((x-1)^2)

So, the derivative of (x+1)/(x-1) using the limit definition is 2 / ((x-1)^2).