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

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Answer 1 · 2017-12-10T12:17:52

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

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

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

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

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