What is the derivative of (2x^2-6x+1^{-8}) ?

The derivative of (2x^2-6x+1^{-8}) is -(8(4x-6))/((2x^2-6x+1)^9)

What is the first derivative of (2x^2-6x+1^{-8}) ?

The first derivative of (2x^2-6x+1^{-8}) is -(8(4x-6))/((2x^2-6x+1)^9)

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

- \frac{8 ( 4 x - 6 )}{( 2 x ^{2} - 6 x + 1 ) ^{9}}

Solution steps

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

Apply the chain rule:

- \frac{8}{( 2 x ^{2} - 6 x + 1 ) ^{9}} \frac{d}{d x} (2 x^{2} - 6 x + 1)

= - \frac{8}{( 2 x ^{2} - 6 x + 1 ) ^{9}} \frac{d}{d x} (2 x^{2} - 6 x + 1)

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

= - \frac{8}{( 2 x ^{2} - 6 x + 1 ) ^{9}} (4 x - 6)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{8 ( 4 x - 6 )}{( 2 x ^{2} - 6 x + 1 ) ^{9}}

\int \frac{1}{2 x - 1} d x

y^{^{''}} + 16 y = 8 tan (4 x)

\frac{2 y ^{^{'}}}{y} = x

\int \frac{2 x ^{3} - 4 x ^{2} - x - 3}{x ^{2} - 2 x - 3} d x

y^{^{''}} + 4 y = - 3 x^{2} + 2 x + 3

What is the derivative of (2x^2-6x+1^{-8}) ?
The derivative of (2x^2-6x+1^{-8}) is -(8(4x-6))/((2x^2-6x+1)^9)
What is the first derivative of (2x^2-6x+1^{-8}) ?
The first derivative of (2x^2-6x+1^{-8}) is -(8(4x-6))/((2x^2-6x+1)^9)