What is the derivative of ((-6x))/(sqrt(8-2x)) ?

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

What is the first derivative of ((-6x))/(sqrt(8-2x)) ?

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

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derivative of

\frac{( - 6 x )}{8 - 2 x}

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

Solution steps

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

Simplify

\frac{- 6 x}{8 - 2 x} : - \frac{6 x}{8 - 2 x}

= \frac{d}{d x} (- \frac{6 x}{8 - 2 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 6 \frac{d}{d x} (\frac{x}{8 - 2 x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= - 6 \cdot \frac{\frac{d x}{d x} 8 - 2 x - \frac{d}{d x} ( 8 - 2 x ) x}{( 8 - 2 x ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (8 - 2 x) = - \frac{1}{8 - 2 x}

= - 6 \cdot \frac{1 \cdot 8 - 2 x - ( - \frac{1}{8 - 2 x} ) x}{( 8 - 2 x ) ^{2}}

Simplify

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

= - \frac{3 ( - x + 8 )}{( - x + 4 ) - 2 x + 8}

y^{^{''}} + y = x, y (1) = 0, y^{^{'}} (1) = 1

\int 3 x^{2} d x

\frac{\partial}{\partial y} (x cos (z))

7 x^{2} - 9 x, at x = - 17

f (x) = 2 x^{2} - 5 x + 3, at x = 1

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