derivative of (x+1/(sqrt(x^2+1)))

解答

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

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

求解步骤

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

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (x + 1) = 1

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

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

化简

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

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

\int \frac{1}{4} x^{4} d x

\frac{d}{d x} (e^{- 2 x} - 2 e^{- 2 x} x)

\frac{d}{d t} (- 1 cos (t) + 8 sin (t))

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\int \frac{x ^{5}}{( 3 - x ^{6} ) ^{4}} d x