What is the derivative of y^2+sqrt(y) ?

The derivative of y^2+sqrt(y) is 2y+1/(2sqrt(y))

What is the first derivative of y^2+sqrt(y) ?

The first derivative of y^2+sqrt(y) is 2y+1/(2sqrt(y))

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derivative of

y^{2} + y

2 y + \frac{1}{2 y}

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}{d y} (y)

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

\frac{d}{d y} (y) = \frac{1}{2 y}

= 2 y + \frac{1}{2 y}

x \to 3 - lim (\frac{- x}{x - 3})

ln (A x^{4} + B)

x \to - 3 lim (∣ x - 1 ∣)

\frac{\partial}{\partial x} (- 7 e^{- 7 y} sin (7 x))

x^{2} + y^{2} = 64, - 8, 3

What is the derivative of y^2+sqrt(y) ?
The derivative of y^2+sqrt(y) is 2y+1/(2sqrt(y))
What is the first derivative of y^2+sqrt(y) ?
The first derivative of y^2+sqrt(y) is 2y+1/(2sqrt(y))