What is the derivative of-x^3+x ?

The derivative of-x^3+x is -3x^2+1

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

The first derivative of-x^3+x is -3x^2+1

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\frac{d}{d 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

4 cos (7 x), 4 - 4 cos (7 x), 0, \frac{π}{7}

\frac{( s + 1 )}{s ^{2} + 1}

\frac{9}{x}

x \to 0 - lim (3^{\frac{1}{x}})

\int \frac{x ^{2}}{x x ^{2} - x + 1} d x

What is the derivative of-x^3+x ?
The derivative of-x^3+x is -3x^2+1
What is the first derivative of-x^3+x ?
The first derivative of-x^3+x is -3x^2+1