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What is the derivative of #ln(x^2+1)^(1/2)#?

1 Answer

Feb 25, 2016

#f'(x)=(1)/((x^2+1))#

Explanation:

#f(x)=ln[(x^2+1)^(1/2)]#

Here we need to apply the chain rule successively:

#f'(x)=(1)/((x^2+1)^(1/2))xx[1/cancel(2)(x^2+1)^(-1/2)xxcancel(2)]#

#f'(x)=(1)/((x^2+1)^(1/2)xx(x^2+1)^(1/2))#

#:.f'(x)=(1)/((x^2+1))#

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