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

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Answer 1 · 2018-03-04T00:17:10

#d/dx (ln(sqrt(x^2+1))) = x/(x^2+1)#

Using the chain rule:

#d/dx (ln(sqrt(x^2+1))) = 1/sqrt(x^2+1) d/dx sqrt(x^2+1)#

#d/dx (ln(sqrt(x^2+1))) = 1/sqrt(x^2+1) 1/(2sqrt(x^2+1))d/dx (x^2+1)#

#d/dx (ln(sqrt(x^2+1))) = 1/(2(x^2+1))(2x)#

#d/dx (ln(sqrt(x^2+1))) = x/(x^2+1)#

We can also note that using the properties of logarithms:

#(ln(sqrt(x^2+1))) = 1/2 ln(x^2+1)#

and simplify the passages:

#d/dx (ln(sqrt(x^2+1))) = 1/2 d/dx ln(x^2+1) = 1/2 1/(x^2+1) d/dx (x^2+1) = (1/2)( 1/(x^2+1) )(2x )= x/(x^2+1)#