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# Evaluate the indicated derivative at the point (1,2) and write the equation for the tangent line in slope-intercept form?

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- VamanLv 71 month ago
1/y +1/x=2

Take the derivative.

-1/y^2 dy/dx-1/x^2=0

dy/dx= -y^2/x^2, at x=1, y=2, the slope=-4

The equation is

y= -4x +c

This is the solution.

- az_lenderLv 71 month ago
You have not "indicated" any particular derivative, but I'll assume what's wanted is dy/dx.

1/y + 1/x = 2 =>

1/y = 2 - 1/x = (2x - 1)/x =>

y = x/(2x - 1) =>

dy/dx = [(2x - 1)*1 - (x)(2)] / (2x-1)^2

= -1/(2x - 1)^2

The point (1,2) is not on the given curve, so the meaning of the question is rather mysterious. Maybe they want a tangent from (1,2) TO the curve? But no, that doesn't work either, as any line through (1,2) will cross the given curve.

Either your teacher is a moron or you have miscopied the question.

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