What is the y^'(x)=(3x^2)/(3y^2-7),y(1)=0 ?

The y^'(x)=(3x^2)/(3y^2-7),y(1)=0 is -(3x^2)/7

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y^'(x)=(3x^2)/(3y^2-7),y(1)=0

y^{^{'}} (x) = \frac{3 x ^{2}}{3 y ^{2} - 7}, y (1) = 0

- \frac{3 x ^{2}}{7}

Solution steps

\frac{3 x ^{2}}{3 y ^{2} - 7} = \frac{3 x ^{2}}{3 y ^{2} - 7}

Constant function results in a constant

y (1) = 0

= \frac{3 x ^{2}}{3 \cdot 0 ^{2} - 7}

Simplify

\frac{3 x ^{2}}{3 \cdot 0 ^{2} - 7} : - \frac{3 x ^{2}}{7}

= - \frac{3 x ^{2}}{7}

f (x) = 3 x + 4, (7, 5)

\frac{\partial}{\partial x} (0.1 \cdot y^{2} + 0.05 \cdot x \cdot y)

\int (sec (x)) d x

\frac{d}{d x} (arcsin (6 x))

\frac{d}{d x} (3 arcsinh (tan (x)))

What is the y^'(x)=(3x^2)/(3y^2-7),y(1)=0 ?
The y^'(x)=(3x^2)/(3y^2-7),y(1)=0 is -(3x^2)/7