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

1 Answer
Steve M
Nov 1, 2016

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

Explanation:

I am assuming by # ln(x^2+1)^(1/2) # that you mean
# ln(sqrt(x^2+1)) # rather than # sqrtln(x^2+1) #

By the chain rule, # d/dxf(g(x)) =f'(g(x))g'(x)# or,# dy/dx=dy/(du)(du)/dx#

So # d/dx{ ln(x^2+1)^(1/2) } = d/dx{ 1/2ln(x^2+1) } # (by the power log rule)

# :. d/dx{ ln(x^2+1)^(1/2) } = 1/2d/dx{ ln(x^2+1) } #
# :. d/dx{ ln(x^2+1)^(1/2) } = 1/2{ 1/(x^2+1)d/dx(x^2+1)} # (by the chain rule)
# :. d/dx{ ln(x^2+1)^(1/2) } = 1/2{ 1/(x^2+1)(2x)} #
# :. d/dx{ ln(x^2+1)^(1/2) } = { 1/(x^2+1)(x)} #

Hence,
# :. d/dx{ ln(x^2+1)^(1/2) } = (x)/(x^2+1) #

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