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

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

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

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

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derivative of

f (x) = \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}}

\frac{\partial}{\partial y} (e^{x} tan (y))

\int 4 x^{- 1} - 4^{x} + 10 e^{2 x} - 8 x d x

\frac{d}{d x} ((x - 7)^{2})

\int x^{3} \cdot e^{- x^{2}} d x

x \to \infty lim (arctan (ln (x)))

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