What is derivative of 2xy-4y+5=0 ?

The answer to derivative of 2xy-4y+5=0 is (dy)/(dx)=-y/(x-2)

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derivative of 2xy-4y+5=0

derivative of

2 x y - 4 y + 5 = 0

\frac{d y}{d x} = - \frac{y}{x - 2}

Solution steps

2 x y - 4 y + 5 = 0

Treat

y

y (x)

Differentiate both sides:

2 (x \frac{d y}{d x} + y) - 4 \frac{d y}{d x} = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = - \frac{y}{x - 2}

\frac{d y}{d x} = - \frac{y}{x - 2}

x = cos^{4} (t)

x^{4} + y^{4} = 4 a^{2} x^{3} y^{3}

(d^{2} + d) \cdot y = x^{2} + 2 x + 4

(d^{2} + d) y = x^{2} + 2 x + 4

x = e^{s i n (t^{2} - 4 t)}

What is derivative of 2xy-4y+5=0 ?
The answer to derivative of 2xy-4y+5=0 is (dy)/(dx)=-y/(x-2)