What is the derivative of x^4-x ?

The derivative of x^4-x is 4x^3-1

What is the first derivative of x^4-x ?

The first derivative of x^4-x is 4x^3-1

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\frac{d}{d x} (x^{4} - x)

4 x^{3} - 1

Solution steps

\frac{d}{d x} (x^{4} - x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) - \frac{d x}{d x}

\frac{d}{d x} (x^{4}) = 4 x^{3}

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

= 4 x^{3} - 1

\frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}}

\int 9 sin^{2} (x) cos^{2} (x) d x

\frac{d y}{d x} = \frac{2 x y}{x ^{2} + 1}

e^{2 x} - 1

c cos (t) + t^{2} sin (t)