Question

Using the chain rule to find the dervative??

Answer 1

\[\frac{ d }{ dx } \sqrt{\frac{ 1-x }{ 1+x }}\]

Answer 2

\[\frac{ 1 }{ 2 }\left( \frac{ 1-x }{ 1+x } \right)^{-1/2} \left( \frac{ 1-x }{ 1+x } \right)\left( \frac{ -2 }{ (1+x^2) } \right)\]
Is this correct for the chain rule?

Answer 3

If so, how would I simplify this?

Answer 4

$$

\Large \frac{ d }{ dx } \Large( \frac{ 1-x }{ 1+x })^{\frac{1}{2}}\\
\Large =\frac{1}{2}\Large( \frac{ 1-x }{ 1+x })^{\frac{-1}{2}} \cdot \frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2}\\
\Large =\frac{1}{2}\Large( \frac{ 1-x }{ 1+x })^{\frac{-1}{2}} \cdot \frac{-2}{(1+x)^2}\\
\Large =\frac{1}{2}\Large( \frac{ 1+x }{ 1-x })^{\frac{1}{2}} \cdot \frac{-2}{(1+x)^2}\\
\Large =\frac{1}{2} \frac{ (1+x)^\frac{1}{2} }{ (1-x)^\frac{1}{2} } \cdot \frac{-2}{(1+x)^2}\\

\Large =\frac{-2}{2} \frac{ (1+x)^\frac{1}{2} }{ (1+x)^2 } \cdot \frac{1}{(1-x)^{\frac{1}{2}}}\\

\Large = - \frac{ 1 }{ (1+x)^{2-1/2} } \cdot \frac{1}{(1-x)^{\frac{1}{2}}}\\

\Large = - \frac{ 1 }{ (1+x)^{\frac{3}{2}}} \cdot \frac{1}{(1-x)^{\frac{1}{2}}}\\

\Large = - \frac{ 1 }{ (1+x)^{\frac{3}{2}} ~ (1-x)^{\frac { 1 } { 2 } } } \\
$$