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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x - y)

\int sec (x) tan (x) - 3 e^{x} d x

\int_{0}^{5} \frac{w}{w - 2} d w

x \to 0 lim (3 x cos (x))

\int_{5}^{7} \frac{1}{x ^{2} - 1} d x

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