What is the derivative of 3x^2y+y^3-3x^2-3y^2+2 ?

The derivative of 3x^2y+y^3-3x^2-3y^2+2 is -6x+3y^2(dy)/(dx)+3(x^2(dy)/(dx)+2xy)-6y(dy)/(dx)

What is the first derivative of 3x^2y+y^3-3x^2-3y^2+2 ?

The first derivative of 3x^2y+y^3-3x^2-3y^2+2 is -6x+3y^2(dy)/(dx)+3(x^2(dy)/(dx)+2xy)-6y(dy)/(dx)

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Popular >

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\frac{d}{d x} (2) = 0

= 3 (2 x y + \frac{d y}{d x} x^{2}) + 3 y^{2} \frac{d y}{d x} - 6 x - 6 y \frac{d y}{d x} + 0

Simplify

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

\int \frac{3}{2 x + 7} d x

\int_{1}^{e} (\frac{1}{x}) d x

\frac{d}{d x} (4 e^{- 4 x})

\int 3 (2 t + 5)^{3} d t

\frac{d}{d x} (arctan (ln (2 x + 1)))

What is the derivative of 3x^2y+y^3-3x^2-3y^2+2 ?
The derivative of 3x^2y+y^3-3x^2-3y^2+2 is -6x+3y^2(dy)/(dx)+3(x^2(dy)/(dx)+2xy)-6y(dy)/(dx)
What is the first derivative of 3x^2y+y^3-3x^2-3y^2+2 ?
The first derivative of 3x^2y+y^3-3x^2-3y^2+2 is -6x+3y^2(dy)/(dx)+3(x^2(dy)/(dx)+2xy)-6y(dy)/(dx)