y^'(x)=((9x)/(9y^2-7))^2,y(1)=0

解

y^{^{'}} (x) = (\frac{9 x}{9 y ^{2} - 7})^{2}, y (1) = 0

\frac{81 x ^{2}}{49}

解答ステップ

以下の仮定で

(\frac{9 x}{9 y ^{2} - 7})^{2}

を見つける:

y (1) = 0

(\frac{9 x}{9 y ^{2} - 7})^{2} = (\frac{9 x}{9 y ^{2} - 7})^{2}

\frac{9 x}{9 y ^{2} - 7} : \frac{9 x}{9 \cdot 0 ^{2} - 7}

= (\frac{9 x}{9 \cdot 0 ^{2} - 7})^{2}

拡張

(\frac{9 x}{9 \cdot 0 ^{2} - 7})^{2} : \frac{81 x ^{2}}{49}

= \frac{81 x ^{2}}{49}