What is (\partial)/(\partial y)(x^2+3x-4xy) ?

The answer to (\partial)/(\partial y)(x^2+3x-4xy) is -4x

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(\partial)/(\partial y)(x^2+3x-4xy)

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

- 4 x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (3 x) = 0

\frac{\partial}{\partial y} (4 x y) = 4 x

= 0 + 0 - 4 x

Simplify

= - 4 x

\int - (\frac{1}{x}) d x

\int - \frac{1}{t} d t

\frac{d}{d x} ((sin (3 x))^{2})

f (x) = x x - x^{2}

e^{x}, ln (12)

What is (\partial)/(\partial y)(x^2+3x-4xy) ?
The answer to (\partial)/(\partial y)(x^2+3x-4xy) is -4x