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

1 Answer
A08
Mar 12, 2016

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

Explanation:

Using the chain rule
If a function #f(x)=[g(x)]^n#

Then #f'(x)=n[g(x)]^(n-1)cdot g'(x)#

Given #g(x)=(x^2+1) and n=1/2#
We see that #g'(x)=2x#
#:. text{derivative of }(x^2+1)^(1/2)=1/2[x^2+1]^(1/2-1)cdot 2x#
#=1/ cancel2[x^2+1]^(-1/2)cdot cancel2 x#
or #=x[x^2+1]^(-1/2)#
or #=x/sqrt(x^2+1)#

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