How do you use the quotient rule to differentiate #x / (1-x^2)^(1/2)#?

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Answer 1 · 2017-04-08T05:44:27

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

Quotient rule states if #f(x)=(g(x))/(h(x))#

then #(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

Here in #x/(1-x^2)^(1/2)#, we have #g(x)=x# and #:.(dg)/(dx)=1#

and #h(x)=(1-x^2)^(1/2)# and

#(dh)/(dx)=1/2(1-x^2)^(-1/2)xx(-2x)=-x/(1-x^2)^(1/2)#

Hence #d/(dx)x/(1-x^2)^(1/2)#

= #(1xx(1-x^2)^(1/2)-(-x/(1-x^2)^(1/2))xx x)/((1-x^2)^(1/2))^2#

= #((1-x^2)^(1/2)+x^2/(1-x^2)^(1/2))/(1-x^2)#

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

= #1/(1-x^2)^(3/2)#