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

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

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

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

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\frac{d}{d x} (x^{4} - \frac{1}{3} x^{3} - \frac{3}{2} x^{2})

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

Solution steps

\frac{d}{d x} (x^{4} - \frac{1}{3} x^{3} - \frac{3}{2} x^{2})

Simplify

x^{4} - \frac{1}{3} x^{3} - \frac{3}{2} x^{2} : x^{4} - \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2}

= \frac{d}{d x} (x^{4} - \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) - \frac{d}{d x} (\frac{x ^{3}}{3}) - \frac{d}{d x} (\frac{3 x ^{2}}{2})

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

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

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

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

n = 0 \sum \infty \frac{n - 1}{n + 1}

a re a x^{2} - 4 x + 3, x

d er i v a t i v e (4 t + 2)^{2} (2 t^{2} - 9)^{- 3}

t a y l or 2 e^{x}

\frac{d}{d x} (- \frac{1}{7})

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