What is the derivative of ln((x+1/(x-3))) ?

The derivative of ln((x+1/(x-3))) is -4/((x+1)(x-3))

What is the first derivative of ln((x+1/(x-3))) ?

The first derivative of ln((x+1/(x-3))) is -4/((x+1)(x-3))

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = \frac{5}{x ^{3}}

\int 8 u^{\frac{7}{2}} d u

5^{x}, 0

y = (x^{4} + 2)^{2} (x^{5} + 4)^{4}

\frac{d}{d x} (e^{4} x^{7})

What is the derivative of ln((x+1/(x-3))) ?
The derivative of ln((x+1/(x-3))) is -4/((x+1)(x-3))
What is the first derivative of ln((x+1/(x-3))) ?
The first derivative of ln((x+1/(x-3))) is -4/((x+1)(x-3))