How do you differentiate #ln(x^2+1)^(1/2)#?

1 Answer
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Mar 13, 2016

#frac{"d"}{"d"x}(ln(x^2+1)^{1/2}) = frac{x}{x^2 + 1} * ln(x^2+1)^{-1/2}#

Explanation:

You need to use the chain rule twice. We do this from inside out.

First let #w = x^2 + 1#.

We differentiate #w# w.r.t. #x# using the power rule.

#frac{"d"w}{"d"x} = 2x#

Next, let #v = ln(x^2+1) = ln(w)#

To differentiate #v# w.r.t. #x#, we use the chain rule.

#frac{"d"v}{"d"x} = frac{"d"v}{"d"w} * frac{"d"w}{"d"x}#

#= frac{"d"}{"d"w}(ln(w)) * (2x)#

#= 1/w * (2x)#

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

Finally, we let #u = ln(x^2+1)^{1/2} = sqrt(v)#. #quad frac{"d"u}{"d"x}# is the derivative that we are seeking. We use the chain rule again.

#frac{"d"u}{"d"x} = frac{"d"u}{"d"v} * frac{"d"v}{"d"x}#

#= frac{"d"}{"d"v}(sqrt(v)) * (2x)/(x^2 + 1)#

#= frac{1}{2sqrt(v)} * (2x)/(x^2 + 1)#

#= frac{1}{2sqrt(ln(x^2+1))} * (2x)/(x^2 + 1)#

#= frac{x}{x^2 + 1} * ln(x^2+1)^{-1/2}#

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