What is the d/(dy)(2x^3+6xy^2-3x^2+3y^2) ?

The d/(dy)(2x^3+6xy^2-3x^2+3y^2) is 12xy+6y

What is the first d/(dy)(2x^3+6xy^2-3x^2+3y^2) ?

The first d/(dy)(2x^3+6xy^2-3x^2+3y^2) is 12xy+6y

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

12 x y + 6 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d y} (6 x y^{2}) = 12 x y

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

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

= 0 + 12 x y - 0 + 6 y

Simplify

= 12 x y + 6 y

\frac{d y}{d x} = \frac{1 + 2 y}{1 + x}

1 + 2 t

\frac{d ^{2}}{d x ^{2}} (\frac{1}{1 + x ^{2}})

\int (3 x - \frac{4}{3} x^{\frac{3}{2}}) d x

x \to 1 - lim (arcsin (\frac{1}{x}))

What is the d/(dy)(2x^3+6xy^2-3x^2+3y^2) ?
The d/(dy)(2x^3+6xy^2-3x^2+3y^2) is 12xy+6y
What is the first d/(dy)(2x^3+6xy^2-3x^2+3y^2) ?
The first d/(dy)(2x^3+6xy^2-3x^2+3y^2) is 12xy+6y