find the equation of a tangent of a curve at point(1,3)

Question

x^2.y^3+y^2.x^4+4x^2-14=0,i tried differentiate it n also applying this formula y=mx+b.bt i gt confuse

Answer 1

In x^2.y^3+y^2.x^4+4x^2-14=0
I will assume you are using the "." as a multiplication symbol
Usually we use the * symbol

x^2(3y^2)dy/dx + y^3(2x) + y^2(4x^3) + x^4(2y)dy/dx + 8x = 0

dy/dx(3x^2 y^2 + 2x^4 y) = -2x y^3 - 4x^3 y^2 -8x
dy/dx = (-2x y^3 - 4x^3 y^2 - 8x)/(3x^2 y^2 + 2x^4 y)
which for (1,3) becomes
dy/dx = (-54 - 36 - 8)/(27+ 6) = -98/33

so in y = mx + b , m = -98/33 and the point is (1,3)
3 = (-98/33)(1) + b
b = 197/33

equation: y = (-98/33)x + 197/33

check my arithmetic, was expecting "nicer" numbers

AHHHH, just found the problem.

One of the first things you should do in these kind of problems is to make sure that the given point is actually on the line.
IT IS NOT, so the above work is meaningless

check your typing of the equation

Answer 2

Ok tankz for the clue u have given me am grateful