What is implicit (dy)/(dx),2x^2+y^2-7=0 ?

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

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implicit (dy)/(dx),2x^2+y^2-7=0

implicit derivative

\frac{d y}{d x}, 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 \frac{1}{x ^{2} + 2 x + 26} d x

(- 12, 0), (- 15, - 3)

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

3 x^{\frac{2}{3}} - 2 x

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