What is the derivative of (x-1|x-1|) ?

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

What is the first derivative of (x-1|x-1|) ?

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

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\frac{d}{d x} ((x - 1) ∣ x - 1 ∣)

\frac{2 ( x - 1 ) ^{2}}{∣ x - 1 ∣}

Solution steps

\frac{d}{d x} ((x - 1) ∣ x - 1 ∣)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x - 1, g = ∣ x - 1 ∣

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

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

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

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

Simplify

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

= \frac{2 ( x - 1 ) ^{2}}{∣ x - 1 ∣}

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

\frac{d}{d x} (k x^{x})

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

\frac{d}{d x} (\frac{3 x + 4}{3 x - 4})

y^{2} = x^{2} (x + 1), (3, 6)

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