Question

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

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