What is the derivative of |x-3|-|2x-2|-1xe[-2,2] ?

The derivative of |x-3|-|2x-2|-1xe[-2,2] is (x-3)/(|x-3|)-(2(2x-2))/(|2x-2|)-1e(-2,2)*1

What is the first derivative of |x-3|-|2x-2|-1xe[-2,2] ?

The first derivative of |x-3|-|2x-2|-1xe[-2,2] is (x-3)/(|x-3|)-(2(2x-2))/(|2x-2|)-1e(-2,2)*1

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\frac{d}{d x} (∣ x - 3 ∣ - ∣ 2 x - 2 ∣ - 1 x e [- 2, 2])

\frac{x - 3}{∣ x - 3 ∣} - \frac{2 ( 2 x - 2 )}{∣ 2 x - 2 ∣} - 1 e (- 2, 2) \cdot 1

Solution steps

\frac{d}{d x} (∣ x - 3 ∣ - ∣ 2 x - 2 ∣ - 1 \cdot x e (- 2, 2))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (∣ x - 3 ∣) - \frac{d}{d x} (∣ 2 x - 2 ∣) - \frac{d}{d x} (1 \cdot x e (- 2, 2))

\frac{d}{d x} (∣ x - 3 ∣) = \frac{x - 3}{∣ x - 3 ∣}

\frac{d}{d x} (∣ 2 x - 2 ∣) = \frac{2 ( 2 x - 2 )}{∣ 2 x - 2 ∣}

\frac{d}{d x} (1 \cdot x e (- 2, 2)) = 1 e (- 2, 2) \cdot 1

= \frac{x - 3}{∣ x - 3 ∣} - \frac{2 ( 2 x - 2 )}{∣ 2 x - 2 ∣} - 1 e (- 2, 2) \cdot 1

\frac{d}{d x} (1 + x ln (x y - 5))

y^{^{'''}} + y^{^{'}} = 0

x \to 2 lim (3 x^{2} + 2 x - 1)

\int \frac{1}{x ^{2} + a} d x

\int \frac{y ^{4} + y ^{2} - 1}{y ^{3} + y} d y

What is the derivative of |x-3|-|2x-2|-1xe[-2,2] ?
The derivative of |x-3|-|2x-2|-1xe[-2,2] is (x-3)/(|x-3|)-(2(2x-2))/(|2x-2|)-1e(-2,2)*1
What is the first derivative of |x-3|-|2x-2|-1xe[-2,2] ?
The first derivative of |x-3|-|2x-2|-1xe[-2,2] is (x-3)/(|x-3|)-(2(2x-2))/(|2x-2|)-1e(-2,2)*1