How do you find the derivative of #y= x/sqrt(x^2+1)# ?
1 Answer
Sep 4, 2014
#y'=1/(x^2+1)^(3/2)# Solution :
#y=x/sqrt(x^2+1)# Using Quotient Rule, which is
#y=f/g# , then#y'=(gf'-fg')/(g^2)# similarly following for the given problem, yields
#y'=(sqrt(x^2+1)-x*1/(2sqrt(x^2+1))*(2x))/(sqrt(x^2+1))^2#
#y'=(sqrt(x^2+1)-x^2/(sqrt(x^2+1)))/(x^2+1)#
#y'=((x^2+1)-x^2)/(x^2+1)^(3/2)#
#y'=1/(x^2+1)^(3/2)#
