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

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

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

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

2 A^{2} x

Solution steps

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

Treat

A, B, y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{\partial}{\partial x} (B^{2} y^{2}) = 0

= 2 A^{2} x + 0

Simplify

= 2 A^{2} x

\frac{d}{d x} (\frac{lo g _{10} ( x )}{x})

n = 0 \sum \infty \frac{1}{0. 5 ^{n}}

(12 - 6 y + e^{- 3 x}) d x - 2 d y = 0

\int \frac{x ^{3}}{x ^{4} + 10} d x

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