derivative of (8x^6-125b^9/(2x^2-5b^3))

解答

\frac{d}{d x} (\frac{8 x ^{6} - 125 b ^{9}}{2 x ^{2} - 5 b ^{3}})

16 x^{3} + 20 b^{3} x

求解步骤

\frac{d}{d x} (\frac{8 x ^{6} - 125 b ^{9}}{2 x ^{2} - 5 b ^{3}})

化简

\frac{8 x ^{6} - 125 b ^{9}}{2 x ^{2} - 5 b ^{3}} : 4 x^{4} + 10 b^{3} x^{2} + 25 b^{6}

= \frac{d}{d x} (4 x^{4} + 10 b^{3} x^{2} + 25 b^{6})

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (4 x^{4}) + \frac{d}{d x} (10 b^{3} x^{2}) + \frac{d}{d x} (25 b^{6})

\frac{d}{d x} (4 x^{4}) = 16 x^{3}

\frac{d}{d x} (10 b^{3} x^{2}) = 20 b^{3} x

\frac{d}{d x} (25 b^{6}) = 0

= 16 x^{3} + 20 b^{3} x + 0

化简

= 16 x^{3} + 20 b^{3} x

d er i v a t i v e (2 x + 1)^{5} (x^{3} - x + 1)^{4}

\frac{d}{d x} (e^{x} (x - 5) + 5)

x \frac{d y}{d x} - y = 6 x^{2}

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{t} + e^{- t}

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