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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{3} - 4 x^{2} + 3) = 6 x^{2} - 8 x

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

= \frac{( 6 x ^{2} - 8 x ) x - 1 \cdot ( 2 x ^{3} - 4 x ^{2} + 3 )}{x ^{2}}

Simplify

\frac{( 6 x ^{2} - 8 x ) x - 1 \cdot ( 2 x ^{3} - 4 x ^{2} + 3 )}{x ^{2}} : \frac{4 x ^{3} - 4 x ^{2} - 3}{x ^{2}}

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

\int_{1}^{7} \frac{5}{x} d x

x = y^{2} + 6 y, y = x

n = 0 \sum \infty x^{n + 5}

x \to 4 + lim (\frac{7}{x - 4})

f (x) = 1 + x

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