I have to calculate the derivative of this:
$$f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$$
But I'm stuck. This is the point where I have arrived:
$$f'(x) = \frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$$
How can I simplify? I didn't include all the passages.
You have $\frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$
splitting into 2 fractions gives ;
$\frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}+\frac{x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$
$=\frac1{x(\ln(10))}+\frac{x}{(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$
$ = \frac1{x(\ln(10))}+\frac{10^y}{\ln(10)(\sqrt{5-x^2})}$
Start with the fact that your function
$$f(x) = \log_{10} (x) - \log_{10}(1+\sqrt{5-x^2}).$$
Since the denominator is positive (when it's defined, that is, for $-\sqrt5\le x\le \sqrt5 $), the function is only defined for $0<x\le \sqrt5 $, so you can write it as $$ f(x)=\log_{10}x-\log_{10}(1+\sqrt{5-x^2}) $$ which should simplify the computation very much. Recall that the derivative of $\log_{10}{x}$ is $$ \frac{1}{x\log10} $$ (natural logarithm, write it ln if you prefer) and apply the chain rule.
You will get $$ \frac{1}{\log10}\left( \frac{1}{x}-\frac{1}{1+\sqrt{5-x^2}}\frac{-x}{\sqrt{5-x^2}} \right) $$ Then it's just simplifications, if you really want to do more than that: $$ \frac{1}{\log10}\frac{\sqrt{5-x^2}+5-x^2+x^2}{x\sqrt{5-x^2}(1+\sqrt{5-x^2})} = \frac{1}{\log10}\frac{\sqrt{5-x^2}+5}{x\sqrt{5-x^2}(1+\sqrt{5-x^2})} $$ The derivative doesn't exist at $x= \sqrt5 $.
The first derivative is given by $$f'(x)= \left( \left( 1+\sqrt {5-{x}^{2}} \right) ^{-1}+{\frac {{x}^{2}}{ \left( 1+\sqrt {5-{x}^{2}} \right) ^{2}\sqrt {5-{x}^{2}}}} \right) \left( 1+\sqrt {5-{x}^{2}} \right) {x}^{-1} \left( \ln \left( 10 \right) \right) ^{-1} $$ and can be simplified to $$f'(x)={\frac {\sqrt {5-{x}^{2}}+5}{x\ln \left( 10 \right) \left( 1+\sqrt { 5-{x}^{2}} \right) \sqrt {5-{x}^{2}}}} $$
