What is the derivative of (x^4-3x^3-6x^2-3x+2/x) ?

The derivative of (x^4-3x^3-6x^2-3x+2/x) is (3x^4-6x^3-6x^2-2)/(x^2)

What is the first derivative of (x^4-3x^3-6x^2-3x+2/x) ?

The first derivative of (x^4-3x^3-6x^2-3x+2/x) is (3x^4-6x^3-6x^2-2)/(x^2)

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\frac{d}{d x} (\frac{x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2}{x})

\frac{3 x ^{4} - 6 x ^{3} - 6 x ^{2} - 2}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2 ) x - \frac{d x}{d x} ( x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2 )}{x ^{2}}

\frac{d}{d x} (x^{4} - 3 x^{3} - 6 x^{2} - 3 x + 2) = 4 x^{3} - 9 x^{2} - 12 x - 3

\frac{d x}{d x} = 1

= \frac{( 4 x ^{3} - 9 x ^{2} - 12 x - 3 ) x - 1 \cdot ( x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2 )}{x ^{2}}

Simplify

\frac{( 4 x ^{3} - 9 x ^{2} - 12 x - 3 ) x - 1 \cdot ( x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2 )}{x ^{2}} : \frac{3 x ^{4} - 6 x ^{3} - 6 x ^{2} - 2}{x ^{2}}

= \frac{3 x ^{4} - 6 x ^{3} - 6 x ^{2} - 2}{x ^{2}}

f (x) = \frac{x}{( 2 x - 1 ) ^{6}}, at x = 1

\frac{d y}{d x} + 3 y + 5 = 0

cos (2 x)

\frac{d}{d x} (\frac{1}{7 x + 1})

\frac{48}{- 4 x}

What is the derivative of (x^4-3x^3-6x^2-3x+2/x) ?
The derivative of (x^4-3x^3-6x^2-3x+2/x) is (3x^4-6x^3-6x^2-2)/(x^2)
What is the first derivative of (x^4-3x^3-6x^2-3x+2/x) ?
The first derivative of (x^4-3x^3-6x^2-3x+2/x) is (3x^4-6x^3-6x^2-2)/(x^2)