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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{4}{10} x^{2} d y - 13 d x = 0

\frac{3}{( s - 2 ) ^{2}}

\frac{\partial}{\partial x} (arctan (7 x t))

\int - 2 sin (2 x) d x

\int \frac{sin ( θ )}{θ} d θ

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