展开 (sec^2(x)(x^2+1)^6)/((e^x-cos^2(x))^4)

解答

展开

\frac{sec ^{2} ( x ) ( x ^{2} + 1 ) ^{6}}{( e ^{x} - cos ^{2} ( x ) ) ^{4}}

\frac{x ^{12} sec ^{2} ( x ) + 6 x ^{10} sec ^{2} ( x ) + 15 x ^{8} sec ^{2} ( x ) + 20 x ^{6} sec ^{2} ( x ) + 15 x ^{4} sec ^{2} ( x ) + 6 x ^{2} sec ^{2} ( x ) + sec ^{2} ( x )}{e ^{4 x} - 4 e ^{3 x} cos ^{2} ( x ) + 6 e ^{2 x} cos ^{4} ( x ) - 4 e ^{x} cos ^{6} ( x ) + cos ^{8} ( x )}

求解步骤

\frac{sec ^{2} ( x ) ( x ^{2} + 1 ) ^{6}}{( e ^{x} - cos ^{2} ( x ) ) ^{4}}

(e^{x} - cos^{2} (x))^{4} = e^{4 x} - 4 e^{3 x} cos^{2} (x) + 6 e^{2 x} cos^{4} (x) - 4 e^{x} cos^{6} (x) + cos^{8} (x)

= \frac{sec ^{2} ( x ) ( x ^{2} + 1 ) ^{6}}{e ^{4 x} - 4 e ^{3 x} cos ^{2} ( x ) + 6 e ^{2 x} cos ^{4} ( x ) - 4 e ^{x} cos ^{6} ( x ) + cos ^{8} ( x )}

(x^{2} + 1)^{6} = x^{12} + 6 x^{10} + 15 x^{8} + 20 x^{6} + 15 x^{4} + 6 x^{2} + 1

= \frac{sec ^{2} ( x ) ( x ^{12} + 6 x ^{10} + 15 x ^{8} + 20 x ^{6} + 15 x ^{4} + 6 x ^{2} + 1 )}{e ^{4 x} - 4 e ^{3 x} cos ^{2} ( x ) + 6 e ^{2 x} cos ^{4} ( x ) - 4 e ^{x} cos ^{6} ( x ) + cos ^{8} ( x )}

乘开

sec^{2} (x) (x^{12} + 6 x^{10} + 15 x^{8} + 20 x^{6} + 15 x^{4} + 6 x^{2} + 1) : x^{12} sec^{2} (x) + 6 x^{10} sec^{2} (x) + 15 x^{8} sec^{2} (x) + 20 x^{6} sec^{2} (x) + 15 x^{4} sec^{2} (x) + 6 x^{2} sec^{2} (x) + sec^{2} (x)

= \frac{x ^{12} sec ^{2} ( x ) + 6 x ^{10} sec ^{2} ( x ) + 15 x ^{8} sec ^{2} ( x ) + 20 x ^{6} sec ^{2} ( x ) + 15 x ^{4} sec ^{2} ( x ) + 6 x ^{2} sec ^{2} ( x ) + sec ^{2} ( x )}{e ^{4 x} - 4 e ^{3 x} cos ^{2} ( x ) + 6 e ^{2 x} cos ^{4} ( x ) - 4 e ^{x} cos ^{6} ( x ) + cos ^{8} ( x )}

e x p an d 1 + t^{2} \cdot (1 + t^{2})

s im pl i f y (2 x - y) (2 x + y)

e x p an d d x (- \frac{7}{3 x ^{3}} + x + \frac{5}{x})

e x p an d x (π - x)

\frac{d ^{2}}{d x ^{2}} ((y) (4 sin (x) + 4 cos (y)))