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

The first d/(dy)(2xy^2-3) is 4xy

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d/(dy)(2xy^2-3)

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

4 x y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{d}{d y} (2 x y^{2}) = 4 x y

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

= 4 x y - 0

Simplify

= 4 x y

t^{4}

x \to 0 + lim (\frac{1}{x ^{2}} - \frac{1}{x})

\int 2 cos (y^{2}) d y

e^{x + y} d x + e^{x - y} d y = 0

f (x) = \frac{x ^{2}}{x + 7}