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

The first d/(dy)(y^2+y) is 2y+1

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

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

2 y + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d y}{d y} = 1

= 2 y + 1

\int \frac{3 x}{1 - 9 x ^{4}} d x

\int \frac{\frac{1}{120}}{- P + 120} d P

tan (sin (x))

\frac{d y}{d x} = \frac{( e ^{x} - e ^{- x} )}{( 3 + 4 y )}

y = x - 1, y^{2} = 2 x + 6