\section*{Problem 5 - V01105}

Differentiate the following function

\[
y=x \cdot \sqrt[5]{x \cdot \sqrt[5]{x}}+\sqrt[5]{\frac{1+x}{1-x}}
\]
}

First, both summands are considered individually. For the first summand, we have

\[
x \cdot \sqrt[5]{x \cdot \sqrt[5]{x}}=x \cdot \sqrt[5]{x^{\frac{6}{5}}}=x^{\frac{31}{25}}
\]
with the derivative

\[
(x \cdot \sqrt[5]{x \cdot \sqrt[5]{x}})^{\prime}=\frac{31}{25} x^{\frac{6}{25}}=\frac{31}{25} \sqrt[25]{x^{6}}
\]

The inner derivative of the second summand is \(\frac{2}{(1-x)^{2}}\). We get

\[
\left(\sqrt[5]{\frac{1+x}{1-x}}\right)^{\prime}=\frac{1}{5}\left(\frac{1+x}{1-x}\right)^{-\frac{4}{5}} \cdot \frac{2}{(1-x)^{2}}
\]

Reformulations yield

\[
\begin{aligned}
\frac{1}{5}\left(\frac{1+x}{1-x}\right)^{-\frac{4}{5}} \cdot \frac{2}{(1-x)^{2}} & =\frac{2}{5} \sqrt[5]{\frac{(1-x)^{4}}{(1+x)^{4}} \cdot \frac{1}{(1-x)^{2}}} \\
& =\frac{2}{5} \sqrt[5]{\frac{(1-x)^{4}}{(1+x)^{4}} \frac{1}{(1-x)^{10}}}=\frac{2}{5} \sqrt[5]{\frac{1}{(1+x)^{4}(1-x)^{6}}} \\
& =\frac{2}{5(1-x)} \sqrt[5]{\frac{1}{(1+x)^{4}(1-x)}}=\frac{2}{5(1-x)} \sqrt[5]{\frac{(1+x)}{(1+x)^{5}(1-x)}} \\
& =\frac{2}{5(1-x)(1+x)} \sqrt[5]{\frac{(1+x)}{(1-x)}}
\end{aligned}
\]

The desired derivative term is thus

\[
y^{\prime}=\frac{31}{25} \sqrt[25]{x^{6}}+\frac{2}{5(1-x)(1+x)} \sqrt[5]{\frac{(1+x)}{(1-x)}}
\]