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

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

Popular >

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

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

6 x^{2} + 2 x y^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

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

Simplify

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

\frac{d y}{d x} = \frac{cos ( 14 x )}{e ^{14 y}}

n = 0 \sum \infty 3 (- 0.1)^{n}

\int arctan (u) d u

\int_{- 4}^{4} 4 (16 - x^{2}) d x

2 x^{3} - 5

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