What is implicit 2x^2+y^2-7=0 ?

The answer to implicit 2x^2+y^2-7=0 is (dy}{dx}=-\frac{2x)/y

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implicit 2x^2+y^2-7=0

implicit derivative

2 x^{2} + y^{2} - 7 = 0

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

Solution steps

2 x^{2} + y^{2} - 7 = 0

Treat

y

y (x)

Differentiate both sides:

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

Isolate

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

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

\int_{π}^{2 π} - cos (x) d x

x \to 1 lim (\frac{x - x ^{3}}{x - 1})

\frac{d}{d x} (\frac{d ^{2}}{d x ^{2}} (x^{4}) + cos (x) + 1)

\int 3 x^{3} e^{5 x^{4}} d x

y^{^{''}} + 3 y^{^{'}} - 10 y = 0

What is implicit 2x^2+y^2-7=0 ?
The answer to implicit 2x^2+y^2-7=0 is (dy}{dx}=-\frac{2x)/y