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標題:
A maths..(different calculus)
發問:
y=e^3x(sinx+5cos5x) y=coskx sinx^k
最佳解答:
(1) y = e3x (sin x + 5 cos 5x) y' = e3x d(sin x + 5 cos 5x)/dx + (sin x + 5 cos 5x) d(e3x)/dx = e3x (cos x - 25 sin 5x) + 3e3x (sin x + 5 cos 5x) = e3x (cos x + 3 sin x + 15 cos 5x - 25 sin 5x) (2) If you are meaning: y = cos kx sin xk y' = cos kx d(sin xk)/dx + sin xk d(cos kx)/dx = cos kx cos xk d(xk)/dx - k sin xk sin kx = kxk-1 cos kx cos xk - k sin xk sin kx If you are meaning: y = cos kx sink x y' = cos kx d(sink x)/dx + sink x d(cos kx)/dx = k sink-1 x cos x cos kx - k sin kx sink x = k sink-1 x (cos x cos kx - sin kx sin x) = k sink-1 x cos [(k + 1)x]
A maths..(different calculus)
發問:
y=e^3x(sinx+5cos5x) y=coskx sinx^k
最佳解答:
(1) y = e3x (sin x + 5 cos 5x) y' = e3x d(sin x + 5 cos 5x)/dx + (sin x + 5 cos 5x) d(e3x)/dx = e3x (cos x - 25 sin 5x) + 3e3x (sin x + 5 cos 5x) = e3x (cos x + 3 sin x + 15 cos 5x - 25 sin 5x) (2) If you are meaning: y = cos kx sin xk y' = cos kx d(sin xk)/dx + sin xk d(cos kx)/dx = cos kx cos xk d(xk)/dx - k sin xk sin kx = kxk-1 cos kx cos xk - k sin xk sin kx If you are meaning: y = cos kx sink x y' = cos kx d(sink x)/dx + sink x d(cos kx)/dx = k sink-1 x cos x cos kx - k sin kx sink x = k sink-1 x (cos x cos kx - sin kx sin x) = k sink-1 x cos [(k + 1)x]
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