some derivative equations

[Guest]

1) How do I simplify this furthur?

. . .f(x) = [e^{-x^3}] / x²

. . .f' = [-3x²e^{-x^3}(x) - (e^{-x^3})] / x²

. . .. .= [e^{-x^3}(-3x²x - 1)] / x²

. . .. .= e^{-x^3}((-3x²x - 1)x^-2)

I'm not too sure if I made this clear, but I tried my best with the brackets.

2) I forget how to do this type of question:

Find the equation of the tangent to the curve defined by y = e^x that is perpendicular to the line defined by 3x + y = 1.

I know you have to find the slope of one of the equations (which one?) and I know when it's perpendicular the slope is the negative recipricol of a slope of one of the equations...?

3) Determine the equation of the tangent for the curve defined by y - e^xy = 0 at the point A(0,1).

 

[skeeter]

(1) I don't know what kind of quotient rule you are using ...

f(x) = e^-x³/x²

f'(x) = [x²*-3x²*e^-x³ - e^-x³*2x]/x⁴

f'(x) = e^-x³[-3x⁴ - 2x]/x⁴

f'(x) = -e^-x³[3x³ + 2]/x³



(2)
3x + y = 1
y = -3x + 1 ... this line has slope, m = -3. a line perpendicular to this one will have slope m = 1/3

y = e^x
dy/dx = e^x = 1/3
x = ln(1/3), y = 1/3

y - (1/3) = (1/3)[x - ln(1/3)]


(3) y - e^xy = 0
dy/dx - e^xy[x(dy/dx) + y] = 0
dy/dx - x*e^xy(dy/dx) = y*e^xy
dy/dx[1 - x*e^xy] = y*e^xy
dy/dx = y*e^xy/[1 - x*e^xy]

you finish up.