What is the derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 ?

The derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 is 12x^5-40x^4+40x^3-20x+8

What is the first derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 ?

The first derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 is 12x^5-40x^4+40x^3-20x+8

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\frac{d}{d x} (2 x^{6} - 8 x^{5} + 10 x^{4} - 10 x^{2} + 8 x - 2)

12 x^{5} - 40 x^{4} + 40 x^{3} - 20 x + 8

Solution steps

\frac{d}{d x} (2 x^{6} - 8 x^{5} + 10 x^{4} - 10 x^{2} + 8 x - 2)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{6}) - \frac{d}{d x} (8 x^{5}) + \frac{d}{d x} (10 x^{4}) - \frac{d}{d x} (10 x^{2}) + \frac{d}{d x} (8 x) - \frac{d}{d x} (2)

\frac{d}{d x} (2 x^{6}) = 12 x^{5}

\frac{d}{d x} (8 x^{5}) = 40 x^{4}

\frac{d}{d x} (10 x^{4}) = 40 x^{3}

\frac{d}{d x} (10 x^{2}) = 20 x

\frac{d}{d x} (8 x) = 8

\frac{d}{d x} (2) = 0

= 12 x^{5} - 40 x^{4} + 40 x^{3} - 20 x + 8 - 0

Simplify

= 12 x^{5} - 40 x^{4} + 40 x^{3} - 20 x + 8

\int \frac{x - 4}{x ^{2} + 2 x - 15} d x

x \to \infty lim (3 x - 9 x^{2} + 6 x)

\int_{0}^{1} \frac{x}{1 + x ^{4}} d x

\frac{d y}{d x} = x y + 2 x + y + 2

y = 3 x, [3, 7]

What is the derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 ?
The derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 is 12x^5-40x^4+40x^3-20x+8
What is the first derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 ?
The first derivative of 2x^6-8x^5+10x^4-10x^2+8x-2 is 12x^5-40x^4+40x^3-20x+8