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)`
`y^4(2y - 2x(dy)/(dx)) = (pi*y^2 - pi*x*2y*(dy)/(dx))(y^2 + 4x^2)`
You need to open the brackets such that:
`2y^5 - 2xy^4(dy)/(dx) = pi*y^4 + 4pix^2*y^2 - 2pi*x*y^3*(dy)/(dx) - 8pix^3*y(dy)/(dx)`
You need to isolate the terms that contain `(dy)/(dx)` to the left side such that:
`2pi*x*y^3*(dy)/(dx)+ 8pix^3*y(dy)/(dx) - 2xy^4(dy)/(dx) = pi*y^4 + 4pix^2*y^2 - 2y^5`
Factoring out `(dy)/(dx)` yields:
`(dy)/(dx)(2pi*x*y^3 + 8pix^3*y - 2xy^4) = pi*y^4 + 4pix^2*y^2 - 2y^5`
You need to divide by `(2pi*x*y^3 + 8pix^3*y - 2xy^4)` such that:
`(dy)/(dx) = (pi*y^4 + 4pix^2*y^2 - 2y^5)/(2pi*x*y^3 + 8pix^3*y - 2xy^4)`
You need to evaluate the slope `(dy)/(dx)` at the point (1,2) such that:
`(dy)/(dx)|_(1,2) = (pi*2^4 + 4pi*1^2*2^2 - 2^5)/(2pi*1*2^3 + 8pi*1^3*2 - 2*1*2^4)`
`(dy)/(dx)|_(1,2) = (16pi + 16pi - 32)/(16pi + 16pi - 32)`
`(dy)/(dx)|_(1,2) = (32 pi - 32)/(32 pi - 32)`
`(dy)/(dx)|_(1,2) = 1`
Hence, evaluating the slope of the tangent line to the given curve, at (1,2), yields `(dy)/(dx)|_(1,2) = 1` .
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