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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = (x^{3} - 9 x)^{12}, at x = 3

\frac{9}{t ^{4}}

\int_{- 5}^{5} (4 - ∣ x ∣) d x

y = 8 csc (2 x^{4} - 7 x + 3)

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