# Question #f3d4d

Mar 21, 2017

See below.

#### Explanation:

Calling $y = {y}_{1} + {y}_{2}$ with

${y}_{1} = {\left(\log x\right)}^{x}$

and

${y}_{2} = {\left({\sin}^{-} x\right)}^{\sin} x$

we have

$\log {y}_{1} = x \log \left(\log x\right)$ and

$\frac{{\mathrm{dy}}_{1}}{{y}_{1}} = \log \left(\log x\right) \mathrm{dx} + \frac{\mathrm{dx}}{\log} \left(x\right)$ and

$\frac{{\mathrm{dy}}_{1}}{\mathrm{dx}} = {y}_{1} \left(\log \left(\log \left(x\right)\right) + \frac{1}{\log} \left(x\right)\right) = {\left(\log x\right)}^{x} \left(\log \left(\log \left(x\right)\right) + \frac{1}{\log} \left(x\right)\right)$

The same procedure for ${y}_{2}$

$\log \left({y}_{2}\right) = \sin x \log \left({\sin}^{-} 1 x\right)$ then

$\frac{{\mathrm{dy}}_{2}}{\mathrm{dx}} = {y}_{2} \left(\cos x \log \left({\sin}^{-} 1 x\right) + \frac{\sin x}{\log \left({\sin}^{-} 1 x\right) \sqrt{1 - {x}^{2}}}\right)$

or

$\frac{{\mathrm{dy}}_{2}}{\mathrm{dx}} = {\left({\sin}^{-} x\right)}^{\sin} x \left(\cos x \log \left({\sin}^{-} 1 x\right) + \frac{\sin x}{\log \left({\sin}^{-} 1 x\right) \sqrt{1 - {x}^{2}}}\right)$

and after that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\mathrm{dy}}_{1}}{\mathrm{dx}} + \frac{{\mathrm{dy}}_{2}}{\mathrm{dx}}$