Question

The first derivative of the function
$\left[{\cos }^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right)+x^x\right]$
with respect to$x$ at $x=1$ is

Answer 1

$u={\cos }^{-1}\left(90-\frac{{\cos }^{-1/4}}{2}\right)$
$=\frac{1}{{\sqrt{1-\left(\frac{90{\cos }^{-1/4}}{2}\right)}}^2}\times dx\left(\frac{90-{\cos }^{-1/4}}{2}\right)$$$V=x^4$$
$=\frac{-2}{\sqrt{4-{\left(90-{\cos }^{-1/4}\right)}^2}}\left(\frac{1}{\sqrt{1-x^2}}\right)$
$V=x^4$
$\log V=x\log x$
$\frac{1}{v}\frac{dv}{dx}=x\times \frac{1}{x}+\log x$
$\frac{dv}{dx}=x^4(1+\log x)$
$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}=\frac{-2}{\sqrt{4-{(90-{\cos }^{-1/4})}^2\sqrt{1-x^2}}}+x^4(1+\cos )$