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How do you find the derivative of #y= x/sqrt(x^2+1)# ?

Answer 1

Christopher Agresta

#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)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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