What is (\partial)/(\partial y)(xy^2+bx^2y) ?

The answer to (\partial)/(\partial y)(xy^2+bx^2y) is 2xy+x^2b

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(\partial)/(\partial y)(xy^2+bx^2y)

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

2 x y + x^{2} b

Solution steps

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

Treat

x, b

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= 2 x y + x^{2} b

\frac{d}{d x} (- x y)

y = e^{4 c o s (x)}

\int e^{x} + \frac{1}{x} d x

f (x) = sin (cos (x))

\int \frac{3}{x ^{2} + 2} d x

What is (\partial)/(\partial y)(xy^2+bx^2y) ?
The answer to (\partial)/(\partial y)(xy^2+bx^2y) is 2xy+x^2b