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

Question

1426 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-13T13:24:49

#d/dx ( x/(x+1) ) = 1/((x+1)^2 #

By definition we have:

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

So:

#d/dx ( x/(x+1) ) = lim_(h->0) ( (x+h)/(x+h+1) - x/(x+1)) /h#

#d/dx ( x/(x+1) ) = lim_(h->0) 1/h(( (x+h)(x+1) - x(x+h+1))/((x+1)(x+h+1))) #

#d/dx ( x/(x+1) ) = lim_(h->0) 1/h((cancelx^2+cancel(hx)+cancelx+h - cancelx^2 -cancel(hx) -cancelx)/((x+1)(x+h+1)) )#

#d/dx ( x/(x+1) ) = lim_(h->0) 1/cancel(h)(cancel(h)/((x+1)(x+h+1)) )#

#d/dx ( x/(x+1) ) = lim_(h->0) 1/((x+1)(x+h+1)) #

#d/dx ( x/(x+1) ) = 1/((x+1)^2 #