What is the derivative of 1/4 x^4-x^3 ?

The derivative of 1/4 x^4-x^3 is x^3-3x^2

What is the first derivative of 1/4 x^4-x^3 ?

The first derivative of 1/4 x^4-x^3 is x^3-3x^2

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\frac{d}{d x} (\frac{1}{4} x^{4} - x^{3})

x^{3} - 3 x^{2}

Solution steps

\frac{d}{d x} (\frac{1}{4} x^{4} - x^{3})

Simplify

\frac{1}{4} x^{4} - x^{3} : \frac{x ^{4}}{4} - x^{3}

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

Apply the Sum/Difference Rule:

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

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

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

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

= x^{3} - 3 x^{2}

\int_{2}^{\infty} y e^{- 7 y} d y

\int x \cdot e^{- \frac{x}{4}} d x

\int \frac{6 + t + t ^{2}}{t} d t

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

x^{6}, x^{3}

What is the derivative of 1/4 x^4-x^3 ?
The derivative of 1/4 x^4-x^3 is x^3-3x^2
What is the first derivative of 1/4 x^4-x^3 ?
The first derivative of 1/4 x^4-x^3 is x^3-3x^2