What is the derivative of ((x^3))/3+x ?

The derivative of ((x^3))/3+x is x^2+1

What is the first derivative of ((x^3))/3+x ?

The first derivative of ((x^3))/3+x is x^2+1

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derivative of

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

x^{2} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

= x^{2} + 1

\int_{8}^{11} 2 arccot (x) d x

t^{2} e^{- 2 t} sin (2 t)

\frac{\partial}{\partial x} (2 x + y)

7 y^{^{'}} - 14 y = 0

\int tan (x) d x

What is the derivative of ((x^3))/3+x ?
The derivative of ((x^3))/3+x is x^2+1
What is the first derivative of ((x^3))/3+x ?
The first derivative of ((x^3))/3+x is x^2+1