What is the derivative of x^4-x^3+x ?

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

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

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

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

4 x^{3} - 3 x^{2} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= 4 x^{3} - 3 x^{2} + 1

n = 0 \sum \infty \frac{n !}{4 ^{n}}

\int_{8}^{9} \frac{x}{x - 8} d x

\frac{\partial}{\partial x} (\frac{2 x}{1 + x + 2 y})

f (x) = ln (2 x + 1)

\frac{e ^{x}}{8 x ^{2} + 1}

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