What is (\partial)/(\partial x)(8x^2-4xy-3x^2y) ?

The answer to (\partial)/(\partial x)(8x^2-4xy-3x^2y) is 16x-4y-6yx

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(\partial)/(\partial x)(8x^2-4xy-3x^2y)

\frac{\partial}{\partial x} (8 x^{2} - 4 x y - 3 x^{2} y)

16 x - 4 y - 6 y x

Solution steps

\frac{\partial}{\partial x} (8 x^{2} - 4 x y - 3 x^{2} y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (8 x^{2}) - \frac{\partial}{\partial x} (4 x y) - \frac{\partial}{\partial x} (3 x^{2} y)

\frac{\partial}{\partial x} (8 x^{2}) = 16 x

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

\frac{\partial}{\partial x} (3 x^{2} y) = 6 y x

= 16 x - 4 y - 6 y x

2 ln (cos (t))

x \to \infty lim (arctan (x))

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

- 2 t^{3} δ (t - 4)

y^{^{''}} + 3 y^{^{'}} + y = (2 - 6 x) cos (x) - 9 sin (x)

What is (\partial)/(\partial x)(8x^2-4xy-3x^2y) ?
The answer to (\partial)/(\partial x)(8x^2-4xy-3x^2y) is 16x-4y-6yx