What is the derivative of f(x)=(-9x)/(sqrt(2x-1)) ?

The derivative of f(x)=(-9x)/(sqrt(2x-1)) is -(9(x-1))/((2x-1)sqrt(2x-1))

What is the first derivative of f(x)=(-9x)/(sqrt(2x-1)) ?

The first derivative of f(x)=(-9x)/(sqrt(2x-1)) is -(9(x-1))/((2x-1)sqrt(2x-1))

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derivative of

f (x) = \frac{- 9 x}{2 x - 1}

- \frac{9 ( x - 1 )}{( 2 x - 1 ) 2 x - 1}

Solution steps

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

Simplify

\frac{- 9 x}{2 x - 1} : - \frac{9 x}{2 x - 1}

= \frac{d}{d x} (- \frac{9 x}{2 x - 1})

Take the constant out:

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

= - 9 \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}}

= - 9 \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}

= - 9 \cdot \frac{1 \cdot 2 x - 1 - \frac{1}{2 x - 1} x}{( 2 x - 1 ) ^{2}}

Simplify

- 9 \cdot \frac{1 \cdot 2 x - 1 - \frac{1}{2 x - 1} x}{( 2 x - 1 ) ^{2}} : - \frac{9 ( x - 1 )}{( 2 x - 1 ) 2 x - 1}

= - \frac{9 ( x - 1 )}{( 2 x - 1 ) 2 x - 1}

\int \frac{cos ^{3} ( x )}{1 - sin ( x )} d x

\frac{d}{d x} (se xx)

\int_{- 1}^{- 0.5} 1.5 x^{2} d x

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

x \to 0 lim (\frac{x ^{2} - 4}{x})

What is the derivative of f(x)=(-9x)/(sqrt(2x-1)) ?
The derivative of f(x)=(-9x)/(sqrt(2x-1)) is -(9(x-1))/((2x-1)sqrt(2x-1))
What is the first derivative of f(x)=(-9x)/(sqrt(2x-1)) ?
The first derivative of f(x)=(-9x)/(sqrt(2x-1)) is -(9(x-1))/((2x-1)sqrt(2x-1))