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

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

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

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

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

Solution steps

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

Treat

A, B, y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\frac{\partial}{\partial x} (a (x - b))

\int \frac{2 x + 5}{x} d x

\int_{0}^{16} x 16 - x d x

\int_{0}^{\infty} \frac{x}{x ^{3} + 1} d x

x \to 3 lim (\frac{4 x - 5}{5 x - 1})

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