What is the derivative of x^2+y ?

The derivative of x^2+y is 2x+(dy)/(dx)

What is the first derivative of x^2+y ?

The first derivative of x^2+y is 2x+(dy)/(dx)

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

2 x + \frac{d y}{d x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

= 2 x + \frac{d y}{d x}

\frac{\partial}{\partial x} (arccos (\frac{y}{x}))

y^{^{'}} x^{2} - 18 = 4 x

\frac{d}{d x} (\frac{( 6 x ^{2} + 2 x + 8 )}{x})

\int_{0}^{1} \frac{30}{6 x ^{2} + 7 x + 1} d x

\int (6 x^{2} - 5 x + 3) d x

What is the derivative of x^2+y ?
The derivative of x^2+y is 2x+(dy)/(dx)
What is the first derivative of x^2+y ?
The first derivative of x^2+y is 2x+(dy)/(dx)