What is the derivative of ${ln (x^{2} + 1)}^{\frac{1}{2}}$ $ln(x^2+1)^(1/2)$ ?

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Answer 1 · 2015-05-14T16:38:41

Here, we must resort to a three-part chain rule.

First, we must rename the whole problem.

$u = x^{2} + 1$ $u=x^2+1$ , $w = u^{\frac{1}{2}}$ $w=u^(1/2)$ and $z = ln (w)$ $z=ln(w)$

The derivatives of these three are:

$u'=2x$ , $w'=1/(2*u^(1/2)$ and $z'=(w')/w$

Now, multiplying all the derivatives...

$(dy)/(dx)=z'*w'*u'$

$(dy)/(dx) = (w')/(w)*1/(2u^(1/2))*2x$

$(dy)/(dx) = (1/(2u))*(1/(2u^(1/2)))*2x$

Finally:

$(dy)/(dx) = x/(2(x^2-1)^(1/2)$

What is the derivative of ln(x^2+1)^(1/2)ln(x2+1)12ln(x^2+1)^(1/2)?