What is the derivative of f(x)=ln((x-2)/(x+1)) ?

The derivative of f(x)=ln((x-2)/(x+1)) is 3/((x-2)(x+1))

What is the first derivative of f(x)=ln((x-2)/(x+1)) ?

The first derivative of f(x)=ln((x-2)/(x+1)) is 3/((x-2)(x+1))

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int x (7 - y)

\int \frac{1}{9 - 4 x ^{2}} d x

\int_{0}^{\infty} f (a + x) d x

n = 0 \sum \infty \frac{3}{( 1 + i ) ^{n}}

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

What is the derivative of f(x)=ln((x-2)/(x+1)) ?
The derivative of f(x)=ln((x-2)/(x+1)) is 3/((x-2)(x+1))
What is the first derivative of f(x)=ln((x-2)/(x+1)) ?
The first derivative of f(x)=ln((x-2)/(x+1)) is 3/((x-2)(x+1))