What is the first d/(dy)(x^2y^2) ?

The first d/(dy)(x^2y^2) is 2x^2y

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d/(dy)(x^2y^2)

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

2 x^{2} y

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= x^{2} \cdot 2 y^{2 - 1}

Simplify

= 2 x^{2} y

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

\frac{\partial}{\partial x} (x e^{2 t} - 4 x)

a re a x, x - 2, 0

\frac{\partial}{\partial x} (ln (x^{2} - 4 x y + y^{2}))

x \to 0 lim (\frac{4}{17 + x})