$f(x)=\log{\frac{x}{1+\sqrt{5-x^2}}}$
I have to find the derivative of $f(x)$. Please tell me if my steps are correct.
$\frac{1+\sqrt{5-x^2}}{x}×\frac{1}{\ln 10}×\frac{1+\sqrt{5-x^2}-x×\frac{1}{2\sqrt{5-x^2}}}{(1+\sqrt{5-x^2})^2}×\frac{1}{2\sqrt{5-x^2}}×(-2x)$
I would write $$f(x)=\log(x)-\log(1+\sqrt{5-x^2})$$ then we get $$f'(x)=\frac{1}{x}-\frac{1}{1+\sqrt{5-x^2}}\cdot \frac{1}{2}(5-x^2)^{-1/2}\cdot (-2x)$$ if you mean $$\log(x)=\ln(x)$$ the logarithm to the base $e$