What is (\partial)/(\partial x)(x^4-4x^3y+6x^2y^2) ?

The answer to (\partial)/(\partial x)(x^4-4x^3y+6x^2y^2) is 4x^3-12yx^2+12y^2x

(\partial)/(\partial x)(x^4-4x^3y+6x^2y^2)

\frac{\partial}{\partial x} (x^{4} - 4 x^{3} y + 6 x^{2} y^{2})

4 x^{3} - 12 y x^{2} + 12 y^{2} x

Solution steps

\frac{\partial}{\partial x} (x^{4} - 4 x^{3} y + 6 x^{2} y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{4}) - \frac{\partial}{\partial x} (4 x^{3} y) + \frac{\partial}{\partial x} (6 x^{2} y^{2})

\frac{\partial}{\partial x} (x^{4}) = 4 x^{3}

\frac{\partial}{\partial x} (4 x^{3} y) = 12 y x^{2}

\frac{\partial}{\partial x} (6 x^{2} y^{2}) = 12 y^{2} x

= 4 x^{3} - 12 y x^{2} + 12 y^{2} x

s l o p e - 3 x^{2} - 2 x y + 3 y^{3} = 57, (- 4, 3)

\int \frac{1}{1 - sin ( \frac{1}{2} x )} d x

t^{2} y^{^{''}} + t y^{^{'}} + 4 y = 0

\int 3 x^{3} cos (5 x) d x

x \to 5 lim (x^{2} + 11)

What is (\partial)/(\partial x)(x^4-4x^3y+6x^2y^2) ?
The answer to (\partial)/(\partial x)(x^4-4x^3y+6x^2y^2) is 4x^3-12yx^2+12y^2x