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

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

Popular >

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

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

3 x^{2} + 6 x y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (2 y^{4}) = 0

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

= 3 x^{2} - 0 - 0 + 6 y x

Simplify

= 3 x^{2} + 6 x y

x \to \infty lim (\frac{e ^{1 - l n (x)}}{e ^{1 - l n (x) 2}})

x \to 2 lim (f (0) - g (3))

\int cos (x) cos (k x) d x

\int \frac{sin ( x )}{1 + cos ( x )} d x

\frac{s + 2}{s ^{2} ( s + 3 )}

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