What is the derivative of 1/2 x^2-x+1/6 x^3 ?

The derivative of 1/2 x^2-x+1/6 x^3 is x-1+(x^2)/2

What is the first derivative of 1/2 x^2-x+1/6 x^3 ?

The first derivative of 1/2 x^2-x+1/6 x^3 is x-1+(x^2)/2

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\frac{d}{d x} (\frac{1}{2} x^{2} - x + \frac{1}{6} x^{3})

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

Solution steps

\frac{d}{d x} (\frac{1}{2} x^{2} - x + \frac{1}{6} x^{3})

Simplify

\frac{1}{2} x^{2} - x + \frac{1}{6} x^{3} : \frac{x ^{2}}{2} - x + \frac{x ^{3}}{6}

= \frac{d}{d x} (\frac{x ^{2}}{2} - x + \frac{x ^{3}}{6})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{2}}{2}) - \frac{d x}{d x} + \frac{d}{d x} (\frac{x ^{3}}{6})

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

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

\frac{d}{d x} (\frac{x ^{3}}{6}) = \frac{x ^{2}}{2}

= x - 1 + \frac{x ^{2}}{2}

\int x^{2} \cdot x d x

\int \frac{x}{( x + 1 ) ( x + 2 ) ( x + 3 ) ^{2}} d x

y^{^{'}} + 2 y = e^{t}

f (x) = - 6 x^{2} - 3 x - 5 x^{3}

\int (3 - x) 7^{(3 - x)^{2}} d x

What is the derivative of 1/2 x^2-x+1/6 x^3 ?
The derivative of 1/2 x^2-x+1/6 x^3 is x-1+(x^2)/2
What is the first derivative of 1/2 x^2-x+1/6 x^3 ?
The first derivative of 1/2 x^2-x+1/6 x^3 is x-1+(x^2)/2