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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Expand

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

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

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