# Math Help - tangent to a curve (Derivatives)...

1. ## tangent 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.....

2. Originally Posted by cyberdx16
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
The derivative of y=x+2/x=x+2x^(-1)
Is, (using the power rule)
y'=1-2x^(-2)=1-2/x^2
Thus, at x=1 the slope is,
1-2/(1)^2=1-2=-1