implicit (dy)/(dx),tan(x-y)= y/(1-x^2)

解

暗黙的な導関数

\frac{d y}{d x}, tan (x - y) = \frac{y}{1 - x ^{2}}

\frac{d y}{d x} = \frac{2 x y - sec ^{2} ( x - y ) + 2 x ^{2} sec ^{2} ( x - y ) - x ^{4} sec ^{2} ( x - y )}{- sec ^{2} ( x - y ) + 2 x ^{2} sec ^{2} ( x - y ) - x ^{4} sec ^{2} ( x - y ) - 1 + x ^{2}}

解答ステップ

tan (x - y) = \frac{y}{1 - x ^{2}}

y

を

y (x) として扱う

両側を計算:

sec^{2} (x - y) (1 - \frac{d y}{d x}) = \frac{\frac{d y}{d x} ( 1 - x ^{2} ) + 2 x y}{( 1 - x ^{2} ) ^{2}}

分離する

\frac{d y}{d x} : \frac{d y}{d x} = \frac{2 x y - sec ^{2} ( x - y ) + 2 x ^{2} sec ^{2} ( x - y ) - x ^{4} sec ^{2} ( x - y )}{- sec ^{2} ( x - y ) + 2 x ^{2} sec ^{2} ( x - y ) - x ^{4} sec ^{2} ( x - y ) - 1 + x ^{2}}

\frac{d y}{d x} = \frac{2 x y - sec ^{2} ( x - y ) + 2 x ^{2} sec ^{2} ( x - y ) - x ^{4} sec ^{2} ( x - y )}{- sec ^{2} ( x - y ) + 2 x ^{2} sec ^{2} ( x - y ) - x ^{4} sec ^{2} ( x - y ) - 1 + x ^{2}}