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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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
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