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.....

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- Sep 27th 2006, 02:09 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, 04:39 PMThePerfectHacker