What is the derivative of (8x)/(sqrt(3-4x)) ?

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

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

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

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derivative of

\frac{8 x}{3 - 4 x}

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

Solution steps

\frac{d}{d x} (\frac{8 x}{3 - 4 x})

Take the constant out:

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

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

Apply the Quotient Rule:

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

= 8 \cdot \frac{\frac{d x}{d x} 3 - 4 x - \frac{d}{d x} ( 3 - 4 x ) x}{( 3 - 4 x ) ^{2}}

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

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

= 8 \cdot \frac{1 \cdot 3 - 4 x - ( - \frac{2}{3 - 4 x} ) x}{( 3 - 4 x ) ^{2}}

Simplify

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

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

\int x^{5} cos (x^{6}) d x

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

\int_{0}^{1} - 5 x 5^{x} d x

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

\frac{d}{d x} (\frac{3 x}{1 - x})

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