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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{1}{s ^{2} ( s ^{2} + 1 ) ^{2}}

y = e^{x}, y = e^{2 x}, x = ln (2), x = ln (3)

x \to 0 + lim ((1 + 4 x)^{c o t (x)})

x^{2} + 1

y^{^{'}} = 2 y + 39

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