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

The answer to implicit (dy)/(dx),x+y^2=1 is (dy)/(dx)=-1/(2y)

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

implicit derivative

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

\frac{d y}{d x} = - \frac{1}{2 y}

Solution steps

x + y^{2} = 1

Treat

y

y (x)

Differentiate both sides:

1 + 2 y \frac{d y}{d x} = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = - \frac{1}{2 y}

\frac{d y}{d x} = - \frac{1}{2 y}

x \to 0 - lim (3)

\frac{d}{d θ} (8 cos (θ) - 6 sin (θ))

f (x) = \frac{( - 2 )}{x}

\frac{d}{d x} (4 x^{4} + 3 x^{2})

f (x) = 5 + x^{2}

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