# Find the slope of the curve tan^-1(2x/y)=(πx/(y^2)) at the point (1, 2) π=pi

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You should remember that the slope of the curve, represented by a function, at a point, is evaluated using derivative of the function at the point.

Hence, you should differentiate both sides with respect to x such that:

`(d(tan^-1(2x/y)))/(dx) = (d(pix/(y^2)))/(dx)`

`1/(1 + 4x^2/y^2)*(d(2x/y))/(dx) = (pi*y^2 - pi*x*2y*(dy)/(dx))/(y^4)`

`1/(1 + 4x^2/y^2)*(2y - 2x(dy)/(dx))/y^2 = (pi*y^2 - pi*x*2y*(dy)/(dx))/(y^4)`

`(2y - 2x(dy)/(dx))/(y^2 + 4x^2) = (pi*y^2 - pi*x*2y*(dy)/(dx))/(y^4)`

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