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

The first d/(dy)(2y^2+3x) is 4y

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

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

4 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 4 y + 0

Simplify

= 4 y

2 y = 4 x, y = 5, 2 y + 3 x = 7

\frac{\partial}{\partial x} (\frac{53}{5 + x ^{2} + y ^{2}})

x \to 1 lim (sin (2 x))

\frac{d}{d x} (5 sin (π x))

\frac{d}{d x} (\frac{x ^{3}}{3} - 4 x + 1)