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

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

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

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

3 y^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + 3 y^{2}

Simplify

= 3 y^{2}

\int (x + 2) (x - 5) d x

\frac{d}{d x} (e^{\frac{x ^{4}}{10}})

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

y^{^{''}} + 9 y = 2 sin (2 x), y (0) = 0, y^{^{'}} (0) = - 1

\frac{4 x}{( x ^{2} + 1 )}