What is the derivative of 6x^5-20x^4+20x^3-10x+4 ?

The derivative of 6x^5-20x^4+20x^3-10x+4 is 30x^4-80x^3+60x^2-10

What is the first derivative of 6x^5-20x^4+20x^3-10x+4 ?

The first derivative of 6x^5-20x^4+20x^3-10x+4 is 30x^4-80x^3+60x^2-10

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\frac{d}{d x} (6 x^{5} - 20 x^{4} + 20 x^{3} - 10 x + 4)

30 x^{4} - 80 x^{3} + 60 x^{2} - 10

Solution steps

\frac{d}{d x} (6 x^{5} - 20 x^{4} + 20 x^{3} - 10 x + 4)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (6 x^{5}) - \frac{d}{d x} (20 x^{4}) + \frac{d}{d x} (20 x^{3}) - \frac{d}{d x} (10 x) + \frac{d}{d x} (4)

\frac{d}{d x} (6 x^{5}) = 30 x^{4}

\frac{d}{d x} (20 x^{4}) = 80 x^{3}

\frac{d}{d x} (20 x^{3}) = 60 x^{2}

\frac{d}{d x} (10 x) = 10

\frac{d}{d x} (4) = 0

= 30 x^{4} - 80 x^{3} + 60 x^{2} - 10 + 0

Simplify

= 30 x^{4} - 80 x^{3} + 60 x^{2} - 10

\int_{0}^{1} π (2 - \frac{1}{2} x)^{2} d x

\frac{\partial}{\partial t} (4 cos (t))

x \to 0 lim (\frac{1}{x} - \frac{1}{x})

y = \frac{5}{x ^{2} + 1}, (2, 1)

\int 7 x cos (8 x) d x

What is the derivative of 6x^5-20x^4+20x^3-10x+4 ?
The derivative of 6x^5-20x^4+20x^3-10x+4 is 30x^4-80x^3+60x^2-10
What is the first derivative of 6x^5-20x^4+20x^3-10x+4 ?
The first derivative of 6x^5-20x^4+20x^3-10x+4 is 30x^4-80x^3+60x^2-10