a-maths~!!!

(1)y=ln(1-x)

dy/dx = -1/(1-x)

(2)y=e-3x

dy/dx = -3e^(-3x)

(3) y= arcsinx2

dy/dx = 1/\/(1-x^2) *(2x) = 2x/\/(1-x^2)

(4)y=arctanex

dy/dx = 1/\/(1+x^2) *(e^x) = e^x/\/(1+x^2)


(5)y=a√(x+1)

lny = √(x+1)lna

1/y*dy/dx = 1/2\/(x+1)*lna

dy/dx = a^√(x+1) lna/2\/(x+1)

2008-08-03 01:17:16 補充：
等我寫下and證下某d公式..

d(e^x)/dx= e^x

y=lnx >> x=e^y

dx/dy = e^y , dy/dx = 1/e^lnx = 1/ x as b= e^lnb

y = arcsinx

x=siny

dx/dy = cosy = &#92;/(1-sin^2 y) = &#92;/(1-x^2) << cosy>0 as -pi/2











2008-08-03 01:17:20 補充：
y = a^x ,a is constant

lny = xlna

1/y*dy/dx = lna

dy/dx = a^x lna

important formula : dy/dx = 1/(dx/dy)

as dy/dx = dy/du*du/dx

1 = dy/dy=dy/dx*dx/dy

dy/dx= 1/(dx/dy)

2008-08-03 01:18:10 補充：
dx/dy = cosy = &#92;/(1-sin^2 y) = &#92;/(1-x^2) << cosy>0 as -pi/2

hkbillionaire 的concept錯左

y= arcsinx2

Let u = x^2
du/dx = 2x

arcsin(u)
dy/dx = dy/du * du/dx
= 1/sqrt(1-u^2) * 2x
= 2x/sqrt(1-x^4)

y = arctan(e^x)

Let u = e^x
du/dx = e^x

=>
y = arctan(u)
dy/dx = dy/du * du/dx
= 1/(u^2+ 1)*e^x
= e^x / (e^2x + 1)