((x^2-1)/(x^2+1))^'

解答

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

\frac{4 x}{( x ^{2} + 1 ) ^{2}}

求解步骤

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

(x^{2} - 1)^{'} = 2 x

(x^{2} + 1)^{'} = 2 x

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

乘开

2 x (x^{2} + 1) - 2 x (x^{2} - 1) : 4 x

= \frac{4 x}{( x ^{2} + 1 ) ^{2}}

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d er i v a t i v e 3 e^{- x} + e^{4 x}

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