What is implicit (dy)/(dx),9x^2+y^2=1 ?

The answer to implicit (dy)/(dx),9x^2+y^2=1 is (dy}{dx}=-\frac{9x)/y

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implicit (dy)/(dx),9x^2+y^2=1

implicit derivative

\frac{d y}{d x}, 9 x^{2} + y^{2} = 1

\frac{d y}{d x} = - \frac{9 x}{y}

Solution steps

9 x^{2} + y^{2} = 1

Treat

y

y (x)

Differentiate both sides:

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

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = - \frac{9 x}{y}

\frac{d y}{d x} = - \frac{9 x}{y}

x \to - \infty lim (x^{2})

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 12 x^{2}

\int \frac{( 2 x + 11 )}{x ^{2} + x - 6} d x

\int \frac{1}{x 49 + x ^{2}} d x

f (x) = x (5 e^{x} - 4 x^{4})

What is implicit (dy)/(dx),9x^2+y^2=1 ?
The answer to implicit (dy)/(dx),9x^2+y^2=1 is (dy}{dx}=-\frac{9x)/y