By the chain rule with u = 1 + x 2 {\displaystyle u=1+x^{2}} , we have
d d x ( 1 + x 2 ) 50 = d y d x = d y d u ⋅ d u d x = d d u u 50 ⋅ d d x ( 1 + x 2 ) = 50 u 49 ⋅ ( 2 x ) = 50 ( 1 + x 2 ) 49 ( 2 x ) . {\displaystyle {\frac {d}{dx}}(1+x^{2})^{50}={\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}={\frac {d}{du}}u^{50}\cdot {\frac {d}{dx}}(1+x^{2})=50u^{49}\cdot (2x)=50(1+x^{2})^{49}(2x).}
Therefore, y ′ = 50 ( 1 + x 2 ) 49 ( 2 x ) {\displaystyle \color {blue}y'=50(1+x^{2})^{49}(2x)}
, we have
