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

The first d/(dy)(x^2y^3) is 3x^2y^2

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d/(dy)(x^2y^3)

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

3 x^{2} y^{2}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Power Rule:

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

= x^{2} \cdot 3 y^{3 - 1}

Simplify

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

\int - x e^{x^{2}} d x

x = sec (\frac{1}{y})

\int - 4 x^{3} d x

\int_{0}^{1} (6 x^{2} - 1) e^{x} d x

x \to - 1 lim (\frac{1}{x ^{2}} + 3 x - 1)