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

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

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

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

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derivative of

f (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}

n = 0 \sum \infty \frac{1}{e ^{n^{2}}}

\int_{0}^{4} e^{x} sin (x) d x

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

\int \frac{100}{w ^{2} 25 - w ^{2}} d w

x y^{^{'}} - 2 y = 3 x^{2}

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