What is the solution for derivative of y^2=x+1 ?

The solution for derivative of y^2=x+1 is y=sqrt((x^2)/2+x+c_{1)},y=-sqrt((x^2)/2+x+c_{1)}

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derivative of y^2=x+1

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

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

Solution steps

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y^{2} = \int x + 1 d x

\int x + 1 d x = \frac{x ^{2}}{2} + x + c_{1}

y^{2} = \frac{x ^{2}}{2} + x + c_{1}

Isolate

y : y = \frac{x ^{2}}{2} + x + c_{1}, y = - \frac{x ^{2}}{2} + x + c_{1}

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

\frac{d y}{d x} = - \frac{y}{1200 + 10 x}

\frac{d x}{d t} = 7 x t^{5}

\frac{d y}{d x} + (\frac{( 1 - y ^{2} )}{( 1 - x ^{2} )}) = 0

y y^{^{'}} - 2 e x = 0, y (0) = 3

x \cdot \frac{d y}{d x} - y = 6 x^{4} y

What is the solution for derivative of y^2=x+1 ?
The solution for derivative of y^2=x+1 is y=sqrt((x^2)/2+x+c_{1)},y=-sqrt((x^2)/2+x+c_{1)}