y = ( ln ( 1 − x 2 ) ) 1 3 {\displaystyle y=(\ln(1-x^{2}))^{\frac {1}{3}}}
Let u = ln ( 1 − x 2 ) {\displaystyle u=\ln(1-x^{2})} and y = f ( u ) = u 1 3 {\displaystyle y=f(u)=u^{\frac {1}{3}}}
Using chain rule, d y d x = d y d u ⋅ d u d x {\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}}
d y d x = 1 3 u − 2 3 ⋅ d u d x = 1 3 ( ln ( 1 − x 2 ) ) − 2 3 ⋅ d u d x {\displaystyle {\frac {dy}{dx}}={\frac {1}{3}}u^{-{\frac {2}{3}}}\cdot {\frac {du}{dx}}={\frac {1}{3}}(\ln(1-x^{2}))^{-{\frac {2}{3}}}\cdot {\frac {du}{dx}}}
for d u d x {\displaystyle {\frac {du}{dx}}} , we can use chain rule again.
Let v = 1 − x 2 {\displaystyle v=1-x^{2}} , u = ln v {\displaystyle u=\ln v}
then d u d x = d u d v ⋅ d v d x {\displaystyle {\frac {du}{dx}}={\frac {du}{dv}}\cdot {\frac {dv}{dx}}}
d u d x = 1 v ⋅ ( − 2 x ) = 1 1 − x 2 ⋅ ( − 2 x ) {\displaystyle {\frac {du}{dx}}={\frac {1}{v}}\cdot (-2x)={\frac {1}{1-x^{2}}}\cdot (-2x)}
Then substitute back, we have d y d x = 1 3 ( ln ( 1 − x 2 ) ) − 2 3 ⋅ d u d x = 1 3 ( ln ( 1 − x 2 ) ) − 2 3 ⋅ 1 1 − x 2 ⋅ ( − 2 x ) {\displaystyle {\frac {dy}{dx}}={\frac {1}{3}}(\ln(1-x^{2}))^{-{\frac {2}{3}}}\cdot {\frac {du}{dx}}={\frac {1}{3}}(\ln(1-x^{2}))^{-{\frac {2}{3}}}\cdot {\frac {1}{1-x^{2}}}\cdot (-2x)}
answer: 1 3 ( ln ( 1 − x 2 ) ) − 2 3 ⋅ − 2 x 1 − x 2 {\displaystyle \color {blue}{\frac {1}{3}}(\ln(1-x^{2}))^{-{\frac {2}{3}}}\cdot {\frac {-2x}{1-x^{2}}}}
and 
, we can use chain rule again.
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