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

The answer to (\partial)/(\partial x)(4x^2+2xy-8y^2) is 8x+2y

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

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

8 x + 2 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (8 y^{2}) = 0

= 8 x + 2 y - 0

Simplify

= 8 x + 2 y

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

\frac{\partial}{\partial x} (tan (4 + 2 x^{2} y^{4} z^{2}))

x \to 0 lim (\frac{sin ^{4} ( 2 x )}{4 x ^{4}})

\frac{d v}{d t} = 0.64 - \frac{v}{2940} \cdot 8, v (0) = 0

\frac{d x}{d t} = 8 x t^{5}

What is (\partial)/(\partial x)(4x^2+2xy-8y^2) ?
The answer to (\partial)/(\partial x)(4x^2+2xy-8y^2) is 8x+2y