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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derivative of

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 )}

\int 1 + 2 x^{4} (8 x^{3}) d x

y^{^{'}} = \frac{x}{y}

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

\frac{\partial}{\partial x} (x ln (z^{4} + x^{7}))

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

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))