Maths Concours Médecine Marrakech 2009-2010

ceci est un test

[latex] E=mc^2 f(x)=\sqrt{x^{3}+2 x^{2}+3}-a x \sqrt{x+b} [/latex]

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On a $f(x)=\frac{1}{1-x} \ln \left(1+\frac{1}{x}\right)$ $f^{\prime}(x)=\left(\frac{1}{1-x}\right)^{\prime} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x}\left(\ln \left(1+\frac{1}{x}\right)\right)^{\prime}$ $$ =-\frac{(1-x)^{\prime}}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x} \frac{\left(1+\frac{1}{x}\right)^{\prime}}{1+\frac{1}{x}} $$ $$ =\frac{1}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x} \times \frac{-1 / x^{2}}{\frac{x+1}{x}} $$ $$ \begin{aligned} &=\frac{1}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x} \times \frac{-1}{x^{2}} \times \frac{x}{x+1} \\ &=\frac{1}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)-\frac{1}{x\left(1-x^{2}\right)} \end{aligned} $$

On a $f(x)=\frac{1}{1-x} \ln \left(1+\frac{1}{x}\right)$
$f^{\prime}(x)=\left(\frac{1}{1-x}\right)^{\prime} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x}\left(\ln \left(1+\frac{1}{x}\right)\right)^{\prime}$
$$
=-\frac{(1-x)^{\prime}}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x} \frac{\left(1+\frac{1}{x}\right)^{\prime}}{1+\frac{1}{x}}
$$
$$
=\frac{1}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x} \times \frac{-1 / x^{2}}{\frac{x+1}{x}}
$$
$$
\begin{aligned}
&=\frac{1}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)+\frac{1}{1-x} \times \frac{-1}{x^{2}} \times \frac{x}{x+1} \
&=\frac{1}{(1-x)^{2}} \ln \left(1+\frac{1}{x}\right)-\frac{1}{x\left(1-x^{2}\right)}
\end{aligned}
$$