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

The first d/(dy)(x^2-3y^2) is -6y

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

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

- 6 y

Solution steps

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

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} (3 y^{2})

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

\frac{d}{d y} (3 y^{2}) = 6 y

= 0 - 6 y

Simplify

= - 6 y

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