求下列函數的導數

對於y=Cos3x，從基本原理求dy/dx
Sol
dy/dx
=lim(t->0)_[y(x)－y(x+t)]/t
=lim(t->0)_[Cos(3x+3t)－Cos(3x)]/t
=lim(t->0)_[Cos(3x)Cos(3t)－Sin(3x)Sin(3t)－Cos(3x)]/t
=lim(t->0)_｛Cos(3x)[Cos(3t)－1]－Sin(3x)Sin(3t)｝/t
=lim(t->0)_Cos(3x)[Cos(3t)－1]/t－lim(t->0)_Sin(3x)Sin(3t)/t
=Cos(3x)*lim(t->0)_[Cos(3t)－1]/t－Sin(3x)*lim(t->0)_Sin(3t)/t
=3Cos(3x)*lim(t->0)_[Cos(3t)－1]/(3t)－3Sin(3x)*lim(t->0)_Sin(3t)/(3t)
=3Cos(3x)*lim(t->0)_(Cost－1)/t－3Sin(3x)*lim(t->0)_Sint/t
=3Cos(3x)*0－3Sin(3x)*1
=－3Sin(3x)

你們不回答嗎 ?.?

dy/dx
= lim( △x→ ∞ ) (cos 3(x + △x) - cos 3x) / △x
= lim( △x→ ∞ ) - 2 sin (3x + 3△x + 3x)/2 sin (3x + 3△x - 3x)/2 / △x
= lim( △x→ ∞ ) - 2 sin (3x + 1.5△x) sin 1.5△x / △x
= lim( △x→ ∞ ) - 2(1.5) sin (3x + 1.5△x) [sin 1.5△x / (1.5△x)]
= - 3 sin 3x [1]
= - 3 sin 3x

問 「從基本原理求」 就是 不讓你用三角函數的微分法～

但你需要用到 三角函數的複角公式～

2014-03-27 00:50:00 補充：
好呀好呀~

我咁多年來都唔鍾意用 sum to product formula

所以就用d 麻煩方法做~

嘻嘻嘻~

2014-03-27 10:40:48 補充：
嗯嗯嗯～

我想了一晚之後，認為不應該由於自己不喜歡用 sum to product 所以就教人地跟我用個咁麻煩的方法～

所以我都係把答案搬過來好一點～

哈哈哈～

2014-03-27 10:41:39 補充：
dy/dx

= lim( h → 0 ) { cos[3(x + h)] - cos(3x) } / h


Consider
　cos[3(x + h)] - cos(3x)
= cos(3x + 3h) - cos(3x)
= cos(3x)cos(3h) - sin(3x)sin(3h) - cos(3x)
= cos(3x)[ cos(3h) - 1 ] - sin(3x)sin(3h)
= -cos(3x)[ 1 - cos(3h) ] - sin(3x)sin(3h)

2014-03-27 10:42:01 補充：
= -cos(3x)[ 1 - cos(3h) ][ 1 + cos(3h) ]/[ 1 + cos(3h) ] - sin(3x)sin(3h)
= -cos(3x)[ 1 - cos²(3h) ]/[ 1 + cos(3h) ] - sin(3x)sin(3h)
= -cos(3x)sin²(3h)/[ 1 + cos(3h) ] - sin(3x)sin(3h)


Therefore,

dy/dx

= lim( h → 0 ) { cos[3(x + h)] - cos(3x) } / h

2014-03-27 10:42:17 補充：
= lim( h → 0 ) { -cos(3x)sin²(3h)/[ 1 + cos(3h) ] - sin(3x)sin(3h) } / h

= lim( h → 0 ) -cos(3x)sin²(3h)/{ h * [ 1 + cos(3h) ] }
- lim( h → 0 ) sin(3x)sin(3h) / h

= lim( h → 0 ) -cos(3x)sin(3h)[sin(3h)/h]/[ 1 + cos(3h) ]
- lim( h → 0 ) sin(3x)sin(3h) / h

2014-03-27 10:42:29 補充：
= lim( 3h → 0 ) -3cos(3x)sin(3h)[sin(3h)/(3h)]/[ 1 + cos(3h) ]
- lim( 3h → 0 ) 3sin(3x)sin(3h) / (3h)

= -3cos(3x) lim( 3h → 0 ) sin(3h)[sin(3h)/(3h)]/[ 1 + cos(3h) ]
- 3sin(3x) lim( 3h → 0 ) sin(3h) / (3h)

= -3cos(3x) sin(0)[1]/[ 1 + cos(0) ]
- 3sin(3x) [1]

= 0 - 3sin(3x)

= -3sin(3x)