Asked by Tom
considering that x is the independent variable in this equation:
y + y^3 + 3 = e^y^2 + 3^x * cos(3y) - x
Evaluate dy/dx
I get to:
dy/dx (1+3y) = e^y^2 * 3y^2 * dy/dx + 3^x * ln3 * (-sin(3y)) * 3 *dy/dx -1
Will it give me?
dy/dx(1+3y) =
dy/dx((e^y^2 * 3y^2) +
(3^x * ln3 * (-sin(3y)) * 3 ) -1
dy/dx(1+3y) =
dy/dx((e^y^2 * 3y^2) - (9^xsin(3y)ln3))
-1
1 / ((e^y^2 * 3y^2) - (9^xsin(3y)ln3) - (1+3y))
y + y^3 + 3 = e^y^2 + 3^x * cos(3y) - x
Evaluate dy/dx
I get to:
dy/dx (1+3y) = e^y^2 * 3y^2 * dy/dx + 3^x * ln3 * (-sin(3y)) * 3 *dy/dx -1
Will it give me?
dy/dx(1+3y) =
dy/dx((e^y^2 * 3y^2) +
(3^x * ln3 * (-sin(3y)) * 3 ) -1
dy/dx(1+3y) =
dy/dx((e^y^2 * 3y^2) - (9^xsin(3y)ln3))
-1
1 / ((e^y^2 * 3y^2) - (9^xsin(3y)ln3) - (1+3y))
Answers
Answered by
Tom
last one is supposed to be
dy/dx = 1 / ((e^y^2 * 3y^2) - (9^xsin(3y)ln3) - (1+3y))
dy/dx = 1 / ((e^y^2 * 3y^2) - (9^xsin(3y)ln3) - (1+3y))
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