d/(dt)((y^'(t))^2)

解

\frac{d}{d t} ((y^{^{'}} (t))^{2})

2 t y^{2'} + 2' \frac{d y}{d t} y^{2' - 1} t^{2}

解答ステップ

\frac{d}{d t} ((y^{^{'}} (t))^{2})

簡素化

(y^{^{'}} (t))^{2} : t^{2} y^{2^{'}}

= \frac{d}{d t} (t^{2} y^{2^{^{'}}})

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{2}, g = y^{2^{'}}

= \frac{d}{d t} (t^{2}) y^{2^{^{'}}} + \frac{d}{d t} (y^{2^{^{'}}}) t^{2}

\frac{d}{d t} (t^{2}) = 2 t

\frac{d}{d t} (y^{2^{'}}) = 2^{'} \frac{d y}{d t} y^{2^{'} - 1}

= 2 t y^{2^{^{'}}} + 2^{^{'}} \frac{d y}{d t} y^{2^{^{'}} - 1} t^{2}