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

Mar 13, 2016

$\frac{\text{d"}{"d} x}{\ln {\left({x}^{2} + 1\right)}^{\frac{1}{2}}} = \frac{x}{{x}^{2} + 1} \cdot \ln {\left({x}^{2} + 1\right)}^{- \frac{1}{2}}$

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.

$\frac{\text{d"w}{"d} x}{=} 2 x$

Next, let $v = \ln \left({x}^{2} + 1\right) = \ln \left(w\right)$

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}

$= \frac{\text{d"}{"d} w}{\ln \left(w\right)} \cdot \left(2 x\right)$

$= \frac{1}{w} \cdot \left(2 x\right)$

$= \frac{2 x}{{x}^{2} + 1}$

Finally, we let $u = \ln {\left({x}^{2} + 1\right)}^{\frac{1}{2}} = \sqrt{v}$. 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}

$= \frac{\text{d"}{"d} v}{\sqrt{v}} \cdot \frac{2 x}{{x}^{2} + 1}$

$= \frac{1}{2 \sqrt{v}} \cdot \frac{2 x}{{x}^{2} + 1}$

$= \frac{1}{2 \sqrt{\ln \left({x}^{2} + 1\right)}} \cdot \frac{2 x}{{x}^{2} + 1}$

$= \frac{x}{{x}^{2} + 1} \cdot \ln {\left({x}^{2} + 1\right)}^{- \frac{1}{2}}$