derivative of (2x-e^{2x}^{sin(2x)})

解答

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

(2 cos (2 x) ln (2 x - e^{2 x}) + \frac{sin ( 2 x ) ( 2 - 2 e ^{2 x} )}{2 x - e ^{2 x}}) (2 x - e^{2 x})^{s i n (2 x)}

求解步骤

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

使用指数法则:

a^{b} = e^{b l n (a)}

(2 x - e^{2 x})^{s i n (2 x)} = e^{s i n (2 x) l n (2 x - e^{2 x})}

= \frac{d}{d x} (e^{s i n (2 x) l n (2 x - e^{2 x})})

使用连锁律:

e^{s i n (2 x) l n (2 x - e^{2 x})} \frac{d}{d x} (sin (2 x) ln (2 x - e^{2 x}))

= e^{s i n (2 x) l n (2 x - e^{2 x})} \frac{d}{d x} (sin (2 x) ln (2 x - e^{2 x}))

\frac{d}{d x} (sin (2 x) ln (2 x - e^{2 x})) = 2 cos (2 x) ln (2 x - e^{2 x}) + \frac{sin ( 2 x ) ( 2 - 2 e ^{2 x} )}{2 x - e ^{2 x}}

= e^{s i n (2 x) l n (2 x - e^{2 x})} (2 cos (2 x) ln (2 x - e^{2 x}) + \frac{sin ( 2 x ) ( 2 - 2 e ^{2 x} )}{2 x - e ^{2 x}})

e^{s i n (2 x) l n (2 x - e^{2 x})} = (2 x - e^{2 x})^{s i n (2 x)}

= (2 cos (2 x) ln (2 x - e^{2 x}) + \frac{sin ( 2 x ) ( 2 - 2 e ^{2 x} )}{2 x - e ^{2 x}}) (2 x - e^{2 x})^{s i n (2 x)}

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