How do you differentiate  f(x)=sqrt(ln(x^2+3) using the chain rule.?

Apr 9, 2018

$f ' \left(x\right) = \frac{x {\left(\ln \left({x}^{2} + 3\right)\right)}^{- \frac{1}{2}}}{{x}^{2} + 3} = \frac{x}{\left({x}^{2} + 3\right) {\left(\ln \left({x}^{2} + 3\right)\right)}^{\frac{1}{2}}} = \frac{x}{\left({x}^{2} + 3\right) \sqrt{\ln \left({x}^{2} + 3\right)}}$

Explanation:

We are given:
$y = {\left(\ln \left({x}^{2} + 3\right)\right)}^{\frac{1}{2}}$
$y ' = \frac{1}{2} \cdot {\left(\ln \left({x}^{2} + 3\right)\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left[\ln \left({x}^{2} + 3\right)\right]$
$y ' = {\left(\ln \left({x}^{2} + 3\right)\right)}^{- \frac{1}{2}} / 2 \cdot \frac{d}{\mathrm{dx}} \left[\ln \left({x}^{2} + 3\right)\right]$

$\frac{d}{\mathrm{dx}} \left[\ln \left({x}^{2} + 3\right)\right] = \frac{\frac{d}{\mathrm{dx}} \left[{x}^{2} + 3\right]}{{x}^{2} + 3}$

$\frac{d}{\mathrm{dx}} \left[{x}^{2} + 3\right] = 2 x$

$y ' = {\left(\ln \left({x}^{2} + 3\right)\right)}^{- \frac{1}{2}} / 2 \cdot \frac{2 x}{{x}^{2} + 3} = \frac{x {\left(\ln \left({x}^{2} + 3\right)\right)}^{- \frac{1}{2}}}{{x}^{2} + 3} = \frac{x}{\left({x}^{2} + 3\right) {\left(\ln \left({x}^{2} + 3\right)\right)}^{\frac{1}{2}}} = \frac{x}{\left({x}^{2} + 3\right) \sqrt{\ln \left({x}^{2} + 3\right)}}$