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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (\frac{x}{3}) = \frac{1}{3}

\frac{d}{d x} (3) = 0

= - 2 x + \frac{1}{3} - 0

Simplify

= - 2 x + \frac{1}{3}

\int - 5 x e^{x^{2}} d x

\int cos (7 x) cos (3 x) d x

\int e^{- s x} x^{2} d x

x^{2} \cdot ln (7 x)

\int_{\frac{1}{5}}^{7} 10 x ln (5 x) d x

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