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

The derivative of f(x)=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))

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

The first derivative of f(x)=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))

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derivative of

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

\frac{4 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{2} - 1}{x ^{2} + 1}) = \frac{4 x}{( x ^{2} + 1 ) ^{2}}

= \frac{1}{\frac{x ^{2} - 1}{x ^{2} + 1}} \cdot \frac{4 x}{( x ^{2} + 1 ) ^{2}}

Simplify

\frac{1}{\frac{x ^{2} - 1}{x ^{2} + 1}} \cdot \frac{4 x}{( x ^{2} + 1 ) ^{2}} : \frac{4 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )}

= \frac{4 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )}

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

x^{3} y^{3} - y = x

\frac{d}{d x} (sin (x) + cos (y))

f (x) = \frac{3}{2 x - 4}

\frac{\partial}{\partial x} (x e^{4 y} sin (3 z))

What is the derivative of f(x)=ln((x^2-1)/(x^2+1)) ?
The derivative of f(x)=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))
What is the first derivative of f(x)=ln((x^2-1)/(x^2+1)) ?
The first derivative of f(x)=ln((x^2-1)/(x^2+1)) is (4x)/((x^2-1)(x^2+1))