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

The derivative of (-x^2-4x-1/(4x-3)) is (-4x^2+6x+16)/((4x-3)^2)

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

The first derivative of (-x^2-4x-1/(4x-3)) is (-4x^2+6x+16)/((4x-3)^2)

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

\frac{- 4 x ^{2} + 6 x + 16}{( 4 x - 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(- 2 x - 4) (4 x - 3) - 4 (- x^{2} - 4 x - 1) : - 4 x^{2} + 6 x + 16

= \frac{- 4 x ^{2} + 6 x + 16}{( 4 x - 3 ) ^{2}}

\int e^{x^{3} + 1} x^{2} d x

\frac{d}{d x} (ln (e^{3 x} + e^{- 3 x}))

ln (\frac{x + 3}{x - 1})

\frac{d}{d x} (5 sec (4 x))

\int (3 x^{2} + y^{2}) ln (x^{2} + y^{2}) d x

What is the derivative of (-x^2-4x-1/(4x-3)) ?
The derivative of (-x^2-4x-1/(4x-3)) is (-4x^2+6x+16)/((4x-3)^2)
What is the first derivative of (-x^2-4x-1/(4x-3)) ?
The first derivative of (-x^2-4x-1/(4x-3)) is (-4x^2+6x+16)/((4x-3)^2)