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

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

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

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

x^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + x^{2}

Simplify

= x^{2}