What is the derivative of 2x^3+6xy^2-3x^2+3y^2 ?

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

What is the first derivative of 2x^3+6xy^2-3x^2+3y^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

cos (3 x), 2.5

1 - t

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

\int sin (x) cos (x) e^{x} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} + 4})

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