What is the derivative of (3x+4)/(sqrt(2x-1)) ?

The derivative of (3x+4)/(sqrt(2x-1)) is (3x-7)/((2x-1)sqrt(2x-1))

What is the first derivative of (3x+4)/(sqrt(2x-1)) ?

The first derivative of (3x+4)/(sqrt(2x-1)) is (3x-7)/((2x-1)sqrt(2x-1))

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derivative of

\frac{3 x + 4}{2 x - 1}

\frac{3 x - 7}{( 2 x - 1 ) 2 x - 1}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (3 x + 4) = 3

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

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

Simplify

\frac{3 2 x - 1 - \frac{1}{2 x - 1} ( 3 x + 4 )}{( 2 x - 1 ) ^{2}} : \frac{3 x - 7}{( 2 x - 1 ) 2 x - 1}

= \frac{3 x - 7}{( 2 x - 1 ) 2 x - 1}

\frac{d}{d x} (4 x^{- \frac{1}{3}})

y (6 y^{2} - x - 1) d x + 2 x d y = 0

\frac{d}{d x} (\frac{sin ( x )}{cos ^{3} ( x )})

f (x) = \frac{2}{x + 5}, at x = 4

n = 1 \sum \infty \frac{6}{n ^{3}}

What is the derivative of (3x+4)/(sqrt(2x-1)) ?
The derivative of (3x+4)/(sqrt(2x-1)) is (3x-7)/((2x-1)sqrt(2x-1))
What is the first derivative of (3x+4)/(sqrt(2x-1)) ?
The first derivative of (3x+4)/(sqrt(2x-1)) is (3x-7)/((2x-1)sqrt(2x-1))