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

The d/(dy)(y^3+xy^2+y) is 3y^2+2xy+1

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

The first d/(dy)(y^3+xy^2+y) is 3y^2+2xy+1

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

3 y^{2} + 2 x y + 1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d y}{d y} = 1

= 3 y^{2} + 2 x y + 1

x \to \frac{2 π}{3} lim (- tan (2 x))

y^{^{'}} + \frac{2}{200 + t} y = 3

(x - 4) (x + 5) (x + 4)

\frac{\partial}{\partial x} (6 y^{7} x^{5} - 5 y^{6} x^{4})

What is the d/(dy)(y^3+xy^2+y) ?
The d/(dy)(y^3+xy^2+y) is 3y^2+2xy+1
What is the first d/(dy)(y^3+xy^2+y) ?
The first d/(dy)(y^3+xy^2+y) is 3y^2+2xy+1