What is the y^'(x)=((9x)/(9y^2-7))^2,y(1)=0 ?

The y^'(x)=((9x)/(9y^2-7))^2,y(1)=0 is (81x^2)/(49)

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y^'(x)=((9x)/(9y^2-7))^2,y(1)=0

y^{^{'}} (x) = (\frac{9 x}{9 y ^{2} - 7})^{2}, y (1) = 0

\frac{81 x ^{2}}{49}

Solution steps

Find

(\frac{9 x}{9 y ^{2} - 7})^{2}

given

y (1) = 0

(\frac{9 x}{9 y ^{2} - 7})^{2} = (\frac{9 x}{9 y ^{2} - 7})^{2}

\frac{9 x}{9 y ^{2} - 7} : \frac{9 x}{9 \cdot 0 ^{2} - 7}

= (\frac{9 x}{9 \cdot 0 ^{2} - 7})^{2}

Expand

(\frac{9 x}{9 \cdot 0 ^{2} - 7})^{2} : \frac{81 x ^{2}}{49}

= \frac{81 x ^{2}}{49}

What is the y^'(x)=((9x)/(9y^2-7))^2,y(1)=0 ?
The y^'(x)=((9x)/(9y^2-7))^2,y(1)=0 is (81x^2)/(49)