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

The derivative of ((x^2-3x)/(sqrt(4x+1))) is (6x^2-4x-3)/((4x+1)sqrt(4x+1))

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

The first derivative of ((x^2-3x)/(sqrt(4x+1))) is (6x^2-4x-3)/((4x+1)sqrt(4x+1))

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{6 x ^{2} - 4 x - 3}{( 4 x + 1 ) 4 x + 1}

x, 0, 3

n = 2 \sum \infty \frac{8}{( n ^{2} - 1 )}

y = \frac{e ^{x}}{x}

\int (4 sin (x) - 1 + 8 x^{- 5}) d x

\int t^{2} - 4 t + 3 d t

What is the derivative of ((x^2-3x)/(sqrt(4x+1))) ?
The derivative of ((x^2-3x)/(sqrt(4x+1))) is (6x^2-4x-3)/((4x+1)sqrt(4x+1))
What is the first derivative of ((x^2-3x)/(sqrt(4x+1))) ?
The first derivative of ((x^2-3x)/(sqrt(4x+1))) is (6x^2-4x-3)/((4x+1)sqrt(4x+1))