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

The derivative of (|x-2|/(x^2)) is (-x^2+6x-8)/(x^3|x-2|)

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

The first derivative of (|x-2|/(x^2)) is (-x^2+6x-8)/(x^3|x-2|)

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

\frac{- x ^{2} + 6 x - 8}{x ^{3} ∣ x - 2 ∣}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{\frac{x - 2}{∣ x - 2 ∣} x ^{2} - 2 x ∣ x - 2 ∣}{( x ^{2} ) ^{2}} : \frac{- x ^{2} + 6 x - 8}{x ^{3} ∣ x - 2 ∣}

= \frac{- x ^{2} + 6 x - 8}{x ^{3} ∣ x - 2 ∣}

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

\int_{- 3}^{5} 6 - y - \frac{( y - 3 ) ^{2}}{4} d y

f (x) = 7 tan^{3} (x)

1 6^{x}

\int + \frac{16}{( 64 x ^{2} + 1 ) ^{2}} d x

What is the derivative of (|x-2|/(x^2)) ?
The derivative of (|x-2|/(x^2)) is (-x^2+6x-8)/(x^3|x-2|)
What is the first derivative of (|x-2|/(x^2)) ?
The first derivative of (|x-2|/(x^2)) is (-x^2+6x-8)/(x^3|x-2|)