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

1 Answer
Dec 10, 2017

Answer:

#f'(x)=x(1-x^2)^(-1)#

Explanation:

If #f(x)=ln(g(x))#
Then #f'(x)=(g'(x))/(g(x))#

#g(x)=(1-x^2)^(-1/2)=(h(x))^n#
#g'(x)=n* h'(x)* h(x)^(n-1)#
#h'(x)=-2x#
#g'(x)=-1/2*-2x*(1-x^2)^(-3/2)=x(1-x^2)^(-3/2)#

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