Maths M2 derivatives form 5

Find dy/dx for the <first principle>
1) Sin(4x)
y=Sin(4x)
dy/dx=lim(t->0)_｛Sin[4(x+t)]－Sin(4x)｝/t
=lim(t->0)_[Sin(4x+4t)－Sin(4x)]/t
=lim(t->0)_[Sin(4x)Cos(4t)+Cos(4x)Sin(4t)－Sin(4x)]/t
=lim(t->0)_｛Sin(4x)[Cos(4t)－1]+Cos(4x)Sin(4t)｝/t
=lim(t->0)_Sin(4x)[Cos(4t)－1]/t+lim(t->0)_[Cos(4x)Sin(4t)]/t
=Sin(4x)lim(t->0)_[Cos(4t)－1]/t+Cos(4x)lim(t->0)_Sin(4t)/t
=4Sin(4x)lim(t->0)_[Cos(4t)－1]/(4t)+4Cos(4x)lim(t->0)_Sin(4t)/(4t)
=4Sin(4x)lim(t->0)_(Cost－1)/t+4Cos(4x)lim(t->0)_Sint/t
=4Sin(4x)*0+4Cos(4x)*1
=4Cos(4x)
2) Sin(6x)
Sol
y=Sin(6x)
dy/dx=lim(t->0)_｛Sin[6(x+t)]-Sin(6x)｝/t
=lim(t->0)_[Sin(6x+6t)－Sin(6x)]/t
=lim(t->0)_[Sin(6x)Cos(6t)+Cos(6x)Sin(6t)－Sin(6x)]/t
=lim(t->0)_｛Sin(6x)[Cos(6t)－1]+Cos(6x)Sin(6t)｝/t
=lim(t->0)_Sin(6x)[Cos(6t)－1]/t+lim(t->0)_[Cos(6x)Sin(6t)]/t
=Sin(6x)lim(t->0)_[Cos(6t)－1]/t+Cos(6x)lim(t->0)_Sin(6t)/t
=6Sin(6x)lim(t->0)_[Cos(6t)－1]/(6t)+6Cos(6x)lim(t->0)_Sin(6t)/(6t)
=6Sin(6x)lim(t->0)_(Cost－1)/t+6Cos(6x)lim(t->0)_Sint/t
=6Sin(6x)*0+6Cos(6x)*1
=6Cos(6x)

Good，有人用 LaTeX。

也可以注意為了分辨 differential operator 和 variable，可考慮把 d/dx 的 d 用 \mathrm{d} 。

https://www.dropbox.com/s/noitkwrxw8qn3ne/20130906PIC01.png

As the URL