How do you differentiate #f(x)= x/sqrt(1-x^2) # using the quotient rule?

1 Answer
Guilherme N.
Dec 30, 2015

We can see this as a product: #f(x)=x*(1-x^2)^(-1/2)# and use the product rule, as well as - for the second term - the chain rule.

Explanation:

  • Chain rule states that for #y=f(x)g(x)#, then #(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#
  • Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

Renaming #u=1-x^2#, we get that the second term becomes #u^(-1/2)# and chain rule turns out to be applicable.

#(dy)/(dx)=1*(1-x^2)^(-1/2)+x(x/(u^(3/2)))#

#(dy)/(dx)=1/(1-x^2)^(1/2)+x^2/((1-x^2)^(3/2))#

#(dy)/(dx)=(1-cancel(x^2)+cancel(x^2))/(1-x^2)^(3/2)=1/(1-x^2)^(3/2)=(1-x^2)^(-3/2)#

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