What is the derivative of (8x^3-2x+9)/x ?

The derivative of (8x^3-2x+9)/x is (16x^3-9)/(x^2)

What is the first derivative of (8x^3-2x+9)/x ?

The first derivative of (8x^3-2x+9)/x is (16x^3-9)/(x^2)

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derivative of

\frac{8 x ^{3} - 2 x + 9}{x}

\frac{16 x ^{3} - 9}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{8 x ^{3} - 2 x + 9}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 8 x ^{3} - 2 x + 9 ) x - \frac{d x}{d x} ( 8 x ^{3} - 2 x + 9 )}{x ^{2}}

\frac{d}{d x} (8 x^{3} - 2 x + 9) = 24 x^{2} - 2

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

= \frac{( 24 x ^{2} - 2 ) x - 1 \cdot ( 8 x ^{3} - 2 x + 9 )}{x ^{2}}

Simplify

\frac{( 24 x ^{2} - 2 ) x - 1 \cdot ( 8 x ^{3} - 2 x + 9 )}{x ^{2}} : \frac{16 x ^{3} - 9}{x ^{2}}

= \frac{16 x ^{3} - 9}{x ^{2}}

\frac{\partial}{\partial u} (\frac{u}{u + v})

(x - 2) (3 x + 3)

f (x) = ln (1 + 2 x)

\frac{d}{d t} (\frac{1}{t + 1})

y - y^{^{'}} (x - 10) = x

What is the derivative of (8x^3-2x+9)/x ?
The derivative of (8x^3-2x+9)/x is (16x^3-9)/(x^2)
What is the first derivative of (8x^3-2x+9)/x ?
The first derivative of (8x^3-2x+9)/x is (16x^3-9)/(x^2)