If f(x) = sqrt{log(x^2+x+1) + sqrt{cosh(2x-3) | vedclass

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(D) Given $f(x) = \sqrt{\log(x^2+x+1) + \sqrt{\cosh(2x-3)}}$.Applying the chain rule, $f'(x) = \frac{1}{2\sqrt{\log(x^2+x+1) + \sqrt{\cosh(2x-3)}}} \c

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$\frac{\sqrt{\cosh(3)} - \sinh(3)}{2(\cosh(3))^{3/4}}$

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(D) Given $f(x) = \sqrt{\log(x^2+x+1) + \sqrt{\cosh(2x-3)}}$.Applying the chain rule, $f'(x) = \frac{1}{2\sqrt{\log(x^2+x+1) + \sqrt{\cosh(2x-3)}}} \cdot \left( \frac{2x+1}{x^2+x+1} + \frac{2\sinh(2x-3)}{2\sqrt{\cosh(2x-3)}} \right)$.At $x=0$, $f'(0) = \frac{1}{2\sqrt{\log(1) + \sqrt{\cosh(-3)}}} \cdot \left( \frac{1}{1} + \frac{\sinh(-3)}{\sqrt{\cosh(-3)}} \right)$.Since $\log(1) = 0$, $\cosh(-3) = \cosh(3)$, and $\sinh(-3) = -\sinh(3)$:$f'(0) = \frac{1 - \frac{\sinh(3)}{\sqrt{\cosh(3)}}}{2\sqrt{\sqrt{\cosh(3)}}} = \frac{\sqrt{\cosh(3)} - \sinh(3)}{2(\cosh(3))^{1/2} \cdot (\cosh(3))^{1/4}} = \frac{\sqrt{\cosh(3)} - \sinh(3)}{2(\cosh(3))^{3/4}}$.

If $f(x) = \sqrt{\log(x^2+x+1) + \sqrt{\cosh(2x-3)}}$, then $f'(0) =$

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