因?yàn)?y =(sin x)^2,
所以 y' =2 sin x (sin x)'
=2 sin x cos x
= sin 2x,
y'' = 2 cos 2x
= 2 sin (2x +π/2),
y''' =4 cos (2x +π/2)
=4 sin (2x +π).
下面用數(shù)學(xué)歸納法證明,
y的n階導(dǎo)數(shù) =2^(n-1) sin [ 2x +(n-1)π ].(n∈N+)
(1) 當(dāng) n=1 時(shí),等式成立.
(2) 假設(shè) n=k 時(shí),等式成立,
即 y的k階導(dǎo)數(shù) =2^(k-1) sin [ 2x +(k-1)π ],
則 y的(k+1) 階導(dǎo)數(shù) = (2^k) cos [ 2x +(k-1)π ]
= (2^k) sin [ 2x +kπ ].
即 當(dāng) n=k+1 時(shí),等式成立.
綜上,y的n階導(dǎo)數(shù) =2^(n-1) sin [ 2x +(n-1)π ].
= = = = = = = = =
高階導(dǎo)數(shù),常用數(shù)學(xué)歸納法,或萊布尼茲公式.