Find the equation of the tangent of the curve y = (8x)/(x-8) at the point (0,0)
We will be using the quotient rule, although the product rule is also usable and can be run through if the student wishes. Firstly, define u = 8x, v = x-8 for simplicity. Then clearly u' = 8, v' = 1, and so by the quotient rule we get y' = -64/(x-8)2. As we wish to find the tangent at the origin, we need the gradient at the point so we evaluate y'(0) = -1. Finally, using the line equation gives us y-y1 = m(x-x1) ==> y-0 = -1(x-0) ==> y = -x is the tangent to the curve at the origin.
TK
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