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

The derivative of (x^2+1)/(|x|-1) is (x^3-2x|x|-x)/(|x|(|x|-1)^2)

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

The first derivative of (x^2+1)/(|x|-1) is (x^3-2x|x|-x)/(|x|(|x|-1)^2)

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derivative of

\frac{x ^{2} + 1}{∣ x ∣ - 1}

\frac{x ^{3} - 2 x ∣ x ∣ - x}{∣ x ∣ ( ∣ x ∣ - 1 ) ^{2}}

Solution steps

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

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

\frac{d}{d x} (∣ x ∣ - 1) = \frac{x}{∣ x ∣}

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

Simplify

\frac{2 x ( ∣ x ∣ - 1 ) - \frac{x}{∣ x ∣} ( x ^{2} + 1 )}{( ∣ x ∣ - 1 ) ^{2}} : \frac{x ^{3} - 2 x ∣ x ∣ - x}{∣ x ∣ ( ∣ x ∣ - 1 ) ^{2}}

= \frac{x ^{3} - 2 x ∣ x ∣ - x}{∣ x ∣ ( ∣ x ∣ - 1 ) ^{2}}

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

y^{^{'}} + 4 y = 5 x

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

\frac{d}{d x} (x^{2} - x - 6)

x \to 2 - lim (\frac{x + 5}{x - 2})

What is the derivative of (x^2+1)/(|x|-1) ?
The derivative of (x^2+1)/(|x|-1) is (x^3-2x|x|-x)/(|x|(|x|-1)^2)
What is the first derivative of (x^2+1)/(|x|-1) ?
The first derivative of (x^2+1)/(|x|-1) is (x^3-2x|x|-x)/(|x|(|x|-1)^2)