What is (\partial)/(\partial y)(ax^2+by^2) ?

The answer to (\partial)/(\partial y)(ax^2+by^2) is 2by

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(\partial)/(\partial y)(ax^2+by^2)

\frac{\partial}{\partial y} (a x^{2} + b y^{2})

2 b y

Solution steps

\frac{\partial}{\partial y} (a x^{2} + b y^{2})

Treat

a, x, b

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (a x^{2}) + \frac{\partial}{\partial y} (b y^{2})

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

\frac{\partial}{\partial y} (b y^{2}) = 2 b y

= 0 + 2 b y

Simplify

= 2 b y

\int x (1 + x)^{\frac{2}{3}} d x

y = cos (x), x = \frac{π}{2}, x = 2 π, y = 0

n = 0 \sum \infty \frac{2}{3} (\frac{1}{3})^{n}

\int \frac{7 - x}{x ^{2} - 4 x + 4} d x

\frac{d}{d x} (cos (w x))

What is (\partial)/(\partial y)(ax^2+by^2) ?
The answer to (\partial)/(\partial y)(ax^2+by^2) is 2by