differentiation

1.) Did you mean Differentiate 3x^2y + xy^2+1 = 0 in respect of x?
If so, the result is 6xy + y^2 + 1 = 0
And do you need to Differentiate the result in respect of y?
If so, the result is 6x + 2y = 0
2.) y(t) = 30e^(0.05t)
Let u = 0.05t
du/dt = 0.05
du = 0.05 dt

dy(t) / du = 30e^(u)
Sub in u and du
dy(t) = 30 e^(0.05t) (0.05dt)
dy(t)/dt = 1.5 e^(0.05t)
6.) y = e^(x^2+2x)
Let u = x^2 + 2x
du/dx = 2x+2
du = (2x + 2) dx

y = e^u
dy/du = e^u
dy = e^u du
Sub in u and du
dy = e^(x^2+2x) (2x+2) dx
dy/dx = (2x+2) e^(x^2+2x)
7.) y = (ln x)^3
Let u = ln x
du/dx = 1/x
du = 1/x dx

y = u^3
dy/du = 3u^2
dy = 3u^2 du
Sub in u and du
dy = 3(ln x)^2 1/x dx
dy/dx = (3/x) (ln x)^2

2008-10-28 20:42:17 補充：
應該用Produce rule即y=uv-->y`=uv`+vu`
3x^2 y+xy^2+1=0
Differentiate in respect of x
d/dx (3x^2 y)+d/dx (xy^2)+d/dx (1) = 0
3 d/dx (x^2 y)+d/dx (xy^2)+0 = 0
3(x^2(dy/dx)+y(2x))+(xy(dy/dx)+(dx/dx)y^2) = 0
(3x^2)(dy/dx)+6xy+(xy)(dy/dx)+(1)(y^2) = 0
(3x^2)(dy/dx)+(xy)(dy/dx)=-6xy-y^2

2008-10-28 20:42:21 補充：
(dy/dx)(3x^2+xy)=-(6xy+y^2)
dy/dx=-(6xy + y^2)/(3x^2+xy)

1.

3 x^2 y+x y^2+1=0
3(2xy+x^2 y')+(y^2+2xyy')=0
y'=-(6xy+y^2)/(3x^2+2xy)

2.

y'(t)=30*0.05e^(0.05t)=1.5e^(0.05t)

6.

y=e^(x^2+2x)
y'=(2x+2)*e^(x^2+2x)=2(x+1)e^(x^2+2x)

7.

y=(ln x)^3
y'=3(ln x)^2*1/x