so i have here

Find an equation for th tanget to the curve y=x+2/x at the point (1,3)

i dont understand why/how the slope of the curve is 1-2/x^2

so what would be the slope of the equ. y=8/x^2+4.....

Printable View

- Sep 27th 2006, 03:09 PMcyberdx16tangent to a curve (Derivatives)...
so i have here

Find an equation for th tanget to the curve y=x+2/x at the point (1,3)

i dont understand why/how the slope of the curve is 1-2/x^2

so what would be the slope of the equ. y=8/x^2+4..... - Sep 27th 2006, 05:39 PMThePerfectHacker