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

The solution for derivative of y^2=x^2 is y=sqrt((x^3)/3+c_{1)},y=-sqrt((x^3)/3+c_{1)}

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derivative of y^2=x^2

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

y = \frac{x ^{3}}{3} + c_{1}, y = - \frac{x ^{3}}{3} + c_{1}

Solution steps

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

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

then

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

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

\int x^{2} d x = \frac{x ^{3}}{3} + c_{1}

y^{2} = \frac{x ^{3}}{3} + c_{1}

Isolate

y : y = \frac{x ^{3}}{3} + c_{1}, y = - \frac{x ^{3}}{3} + c_{1}

y = \frac{x ^{3}}{3} + c_{1}, y = - \frac{x ^{3}}{3} + c_{1}

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

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

\frac{d y}{d x} = 8 x^{6} y - y, (1) = - 2

\frac{d y}{d x} + 50 y = \frac{1}{2}, y (0) = 0

y^{^{'}} = 8 t - 10 t - 2 t \cdot y

What is the solution for derivative of y^2=x^2 ?
The solution for derivative of y^2=x^2 is y=sqrt((x^3)/3+c_{1)},y=-sqrt((x^3)/3+c_{1)}