What is the derivative of sqrt(2x^3-4x-8x+6) ?

The derivative of sqrt(2x^3-4x-8x+6) is (3(x^2-2))/(sqrt(2x^3-12x+6))

What is the first derivative of sqrt(2x^3-4x-8x+6) ?

The first derivative of sqrt(2x^3-4x-8x+6) is (3(x^2-2))/(sqrt(2x^3-12x+6))

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\frac{d}{d x} (2 x^{3} - 4 x - 8 x + 6)

\frac{3 ( x ^{2} - 2 )}{2 x ^{3} - 12 x + 6}

Solution steps

\frac{d}{d x} (2 x^{3} - 4 x - 8 x + 6)

Apply the chain rule:

\frac{1}{2 2 x ^{3} - 4 x - 8 x + 6} \frac{d}{d x} (2 x^{3} - 4 x - 8 x + 6)

= \frac{1}{2 2 x ^{3} - 4 x - 8 x + 6} \frac{d}{d x} (2 x^{3} - 4 x - 8 x + 6)

\frac{d}{d x} (2 x^{3} - 4 x - 8 x + 6) = 6 x^{2} - 12

= \frac{1}{2 2 x ^{3} - 4 x - 8 x + 6} (6 x^{2} - 12)

Simplify

\frac{1}{2 2 x ^{3} - 4 x - 8 x + 6} (6 x^{2} - 12) : \frac{3 ( x ^{2} - 2 )}{2 x ^{3} - 12 x + 6}

= \frac{3 ( x ^{2} - 2 )}{2 x ^{3} - 12 x + 6}

\frac{1}{s ^{2}} (1 - e^{- 5 s})

12 x^{3} + 24 x^{2}

\frac{\partial}{\partial y} (x^{2} - 4 x y + 4 y^{2} + 2)

n = 1 \sum \infty cos (\frac{2}{n})

x \to 1 + lim (\frac{7}{x ^{3} - 1})

What is the derivative of sqrt(2x^3-4x-8x+6) ?
The derivative of sqrt(2x^3-4x-8x+6) is (3(x^2-2))/(sqrt(2x^3-12x+6))
What is the first derivative of sqrt(2x^3-4x-8x+6) ?
The first derivative of sqrt(2x^3-4x-8x+6) is (3(x^2-2))/(sqrt(2x^3-12x+6))