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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x^{2} - 1, - x + 2, x = 0 x = 1

x^{4} + 9 e^{x}

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

\frac{d y}{d t} - \frac{2 t}{1 + t ^{2}} y = 3

tan^{2} (y)

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