What is the derivative of sqrt(x^2+1)√x2+1?

1 Answer
A08
Mar 12, 2016

x/sqrt(x^2+1)x√x2+1

Explanation:

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