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How does one find the equation of the tangent to the curve at y=2loge(x) at (1,0)?

Cheers.

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1 Answer

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The tangent has the following properties:


is a straight line (and hence has the general equation: y = mx+c)
crosses some specified point,
with the same gradient of that specific point


That specified point is (1,0)



Gradient at that point (requires derivative)


Finding the derivative: dy/dx = 2/x
Gradient at x=1 (substitute x=1 into dy/dx): dy/dx = 2




Tangent


Point: (1,0)
Gradient: m = 2 (i.e. the tangent follows the equation: y = 2x + c)


To find c, substitute in (1,0) (i.e.: x=1, y=0)


y = 2x+c
0 = 2(1) + c
0 = 2 + c
c = -2


Therefore, the tangent is: y = 2x - 2

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