What is the derivative of (-7x/(sqrt(2x-1))) ?

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

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

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

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

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

Solution steps

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

Simplify

\frac{- 7 x}{2 x - 1} : - \frac{7 x}{2 x - 1}

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

= - 7 \cdot \frac{\frac{d x}{d x} 2 x - 1 - \frac{d}{d x} ( 2 x - 1 ) x}{( 2 x - 1 ) ^{2}}

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

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

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

Simplify

- 7 \cdot \frac{1 \cdot 2 x - 1 - \frac{1}{2 x - 1} x}{( 2 x - 1 ) ^{2}} : - \frac{7 ( x - 1 )}{( 2 x - 1 ) 2 x - 1}

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

\frac{\partial}{\partial x} (9 x + 16 ∣ y ∣)

\int \frac{1}{( x ( ( 4 x ^{2} - 9 ) ) )} d x

x \to 5 lim (2 x - 7)

\int \frac{1}{π} x cos (n x) d x

x \to - 4 lim (\frac{4}{x + 4})

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