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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{1}^{6} \frac{4}{x ^{2} + 4 x + 4} d x

\int \frac{arcsin ( x )}{1 - x ^{2}} d x

\frac{\partial}{\partial x} (\frac{( 3 x ^{4} - 2 x y )}{y ^{2}})

\int \frac{2 x}{x ^{4} + 4} d x

\frac{d y}{d x} = \frac{6 x}{y}

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