What is (\partial)/(\partial y)((A^2x^2+B^2y^2)^4) ?

The answer to (\partial)/(\partial y)((A^2x^2+B^2y^2)^4) is 8B^2y(A^2x^2+B^2y^2)^3

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(\partial)/(\partial y)((A^2x^2+B^2y^2)^4)

\frac{\partial}{\partial y} ((A^{2} x^{2} + B^{2} y^{2})^{4})

8 B^{2} y (A^{2} x^{2} + B^{2} y^{2})^{3}

Solution steps

\frac{\partial}{\partial y} ((A^{2} x^{2} + B^{2} y^{2})^{4})

Treat

A, B, x

as a constant

Apply the chain rule:

4 (A^{2} x^{2} + B^{2} y^{2})^{3} \frac{\partial}{\partial y} (A^{2} x^{2} + B^{2} y^{2})

= 4 (A^{2} x^{2} + B^{2} y^{2})^{3} \frac{\partial}{\partial y} (A^{2} x^{2} + B^{2} y^{2})

\frac{\partial}{\partial y} (A^{2} x^{2} + B^{2} y^{2}) = 2 B^{2} y

= 4 (A^{2} x^{2} + B^{2} y^{2})^{3} \cdot 2 B^{2} y

Simplify

= 8 B^{2} y (A^{2} x^{2} + B^{2} y^{2})^{3}

\int \frac{e ^{\frac{4}{x}}}{x ^{2}} d x

x \to - \infty lim (xx)

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

\frac{s + 7}{( s + 2 ) ( s - 5 )}

What is (\partial)/(\partial y)((A^2x^2+B^2y^2)^4) ?
The answer to (\partial)/(\partial y)((A^2x^2+B^2y^2)^4) is 8B^2y(A^2x^2+B^2y^2)^3