Hello I would be greatful if someone could check the following, as I do not 100% trust my algrbra:
Find the derivative, simplify as far as possible and give answer in positive exponents.
\(\displaystyle \L\\\begin{array}{l}
w(x) = \sqrt[3]{{\frac{{x - 1}}{{2x + 1}}}} \\
w(x) = \left( {\frac{{x - 1}}{{2x + 1}}} \right)_{}^{1/3} \\
w'(x) = \frac{1}{3}\left( {\frac{{x - 1}}{{2x + 1}}} \right)^{ - 2/3} \left( {\frac{{(2x + 1) - (x - 1)2}}{{(2x + 1)^2 }}} \right) \\
w'(x) = \frac{1}{3}\left( {\frac{{x - 1}}{{2x + 1}}} \right)^{ - 2/3} \left( {\frac{3}{{(2x + 1)^2 }}} \right) \\
w'(x) = \left( {\frac{{(2x + 1)^{2/3} }}{{(2x + 1)^2 (x - 1)^{2/3} }}} \right) \\
w'(x) = \left( {\frac{1}{{(2x + 1)^{4/3} (x - 1)^{2/3} }}} \right) \\
\end{array}\)
Thanks Sophie
Find the derivative, simplify as far as possible and give answer in positive exponents.
\(\displaystyle \L\\\begin{array}{l}
w(x) = \sqrt[3]{{\frac{{x - 1}}{{2x + 1}}}} \\
w(x) = \left( {\frac{{x - 1}}{{2x + 1}}} \right)_{}^{1/3} \\
w'(x) = \frac{1}{3}\left( {\frac{{x - 1}}{{2x + 1}}} \right)^{ - 2/3} \left( {\frac{{(2x + 1) - (x - 1)2}}{{(2x + 1)^2 }}} \right) \\
w'(x) = \frac{1}{3}\left( {\frac{{x - 1}}{{2x + 1}}} \right)^{ - 2/3} \left( {\frac{3}{{(2x + 1)^2 }}} \right) \\
w'(x) = \left( {\frac{{(2x + 1)^{2/3} }}{{(2x + 1)^2 (x - 1)^{2/3} }}} \right) \\
w'(x) = \left( {\frac{1}{{(2x + 1)^{4/3} (x - 1)^{2/3} }}} \right) \\
\end{array}\)
Thanks Sophie