The first derivative of the function f(x) = c | vedclass

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(D) Let $f(x) = \cos^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right) + x^x$. First,simplify the term $g(x) = \cos^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right

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$\frac{3}{4}$

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(D) Let $f(x) = \cos^{-1}\left(\sin \sqrt{\frac{1+x}{2}}ight) + x^x$. First,simplify the term $g(x) = \cos^{-1}\left(\sin \sqrt{\frac{1+x}{2}}ight)$. Using $\sin(	heta) = \cos(\frac{\pi}{2} - 	heta)$,we get $g(x) = \cos^{-1}\left(\cos(\frac{\pi}{2} - \sqrt{\frac{1+x}{2}})ight) = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}}$. Differentiating $g(x)$ with respect to $x$: $g'(x) = -\frac{1}{2\sqrt{\frac{1+x}{2}}} \cdot \frac{1}{2} = -\frac{1}{4\sqrt{\frac{1+x}{2}}}$. At $x=1$,$g'(1) = -\frac{1}{4\sqrt{1}} = -\frac{1}{4}$. Next,let $h(x) = x^x$. Then $\ln(h(x)) = x \ln(x)$. Differentiating with respect to $x$: $\frac{h'(x)}{h(x)} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$. So,$h'(x) = x^x(\ln(x) + 1)$. At $x=1$,$h'(1) = 1^1(\ln(1) + 1) = 1(0 + 1) = 1$. The derivative of $f(x)$ at $x=1$ is $f'(1) = g'(1) + h'(1) = -\frac{1}{4} + 1 = \frac{3}{4}$.

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

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