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

解答

隐函数求导

\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}}

求解步骤

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

将

y

视为

y (x)

对两侧微分:

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}}

整理

\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}}

\int cot (42 x) d x

\frac{d}{d x} (3^{x^{3} - x + 1})

\frac{d}{d x} (arcsec (\frac{x}{a}))

\int_{- \infty}^{0} \frac{1}{5 - 8 x} d x

\frac{d}{d x} (x^{2} sec (\frac{1}{x ^{3}}))