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

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

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

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

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derivative of

f (x) = x^{3} + x

3 x^{2} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 3 x^{2} + 1

2 x, at x = 9

f (x) = \frac{8 x}{x ^{2} - 1}, at x = 3

(\frac{6 x ^{2}}{8 - x})^{3}

x^{- 5}

\int 4 sin (5 x) d x

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