What is the derivative of x^3+y ?

The derivative of x^3+y is 3x^2+(dy)/(dx)

What is the first derivative of x^3+y ?

The first derivative of x^3+y is 3x^2+(dy)/(dx)

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

x \to \infty lim (x \cdot e^{x})

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

(4 sin (t))^{^{'}}

n = 2 \sum \infty \frac{4}{3 ^{n}}

(x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1)

What is the derivative of x^3+y ?
The derivative of x^3+y is 3x^2+(dy)/(dx)
What is the first derivative of x^3+y ?
The first derivative of x^3+y is 3x^2+(dy)/(dx)