derivative f(x)=arctan((1+x)/(1-x))

解答

导数

f (x) = arctan (\frac{1 + x}{1 - x})

\frac{1}{x ^{2} + 1}

求解步骤

\frac{d}{d x} (arctan (\frac{1 + x}{1 - x}))

使用连锁律:

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

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

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

= \frac{1}{( \frac{1 + x}{1 - x} ) ^{2} + 1} \cdot \frac{2}{( 1 - x ) ^{2}}

化简

\frac{1}{( \frac{1 + x}{1 - x} ) ^{2} + 1} \cdot \frac{2}{( 1 - x ) ^{2}} : \frac{1}{x ^{2} + 1}

= \frac{1}{x ^{2} + 1}

\frac{\partial}{\partial x} (1 + x^{3} + y^{4})

\int \frac{3 x + 4}{x + 3} d x

\int_{0}^{1} - 10 (x^{2} + 1) e^{- x} d x

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

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