What is the derivative of x^{2020}+1/3 x^3-4x^2 ?

The derivative of x^{2020}+1/3 x^3-4x^2 is 2020x^{2019}+x^2-8x

What is the first derivative of x^{2020}+1/3 x^3-4x^2 ?

The first derivative of x^{2020}+1/3 x^3-4x^2 is 2020x^{2019}+x^2-8x

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derivative of

x^{2020} + \frac{1}{3} x^{3} - 4 x^{2}

2020 x^{2019} + x^{2} - 8 x

Solution steps

\frac{d}{d x} (x^{2020} + \frac{1}{3} x^{3} - 4 x^{2})

Simplify

x^{2020} + \frac{1}{3} x^{3} - 4 x^{2} : x^{2020} + \frac{x ^{3}}{3} - 4 x^{2}

= \frac{d}{d x} (x^{2020} + \frac{x ^{3}}{3} - 4 x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2020}) + \frac{d}{d x} (\frac{x ^{3}}{3}) - \frac{d}{d x} (4 x^{2})

\frac{d}{d x} (x^{2020}) = 2020 x^{2019}

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

\frac{d}{d x} (4 x^{2}) = 8 x

= 2020 x^{2019} + x^{2} - 8 x

\frac{\partial}{\partial y} (x e^{2 x + 3 y})

\int (\frac{cos ( x )}{sin ^{2} ( x )}) d x

\int u d u

(ln^{3} (x))^{^{'}}

\int (sin (x))^{5} (cos (x))^{3} d x

What is the derivative of x^{2020}+1/3 x^3-4x^2 ?
The derivative of x^{2020}+1/3 x^3-4x^2 is 2020x^{2019}+x^2-8x
What is the first derivative of x^{2020}+1/3 x^3-4x^2 ?
The first derivative of x^{2020}+1/3 x^3-4x^2 is 2020x^{2019}+x^2-8x