What is the derivative of f(x)=|x^3-3x^2+2| ?

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

What is the first derivative of f(x)=|x^3-3x^2+2| ?

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

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derivative of

f (x) = x^{3} - 3 x^{2} + 2

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (x^{3} - 3 x^{2} + 2) = 3 x^{2} - 6 x

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

\frac{d y}{d t} = \frac{( t e ^{t} )}{y 5 + y ^{2}}

\frac{\partial}{\partial x} (c x)

f (x) = \frac{8}{x}, at x = 2

\int \frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} d x

\frac{\partial}{\partial x} (6 x^{\frac{1}{3}} y^{\frac{2}{3}})

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