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

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

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

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

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derivative of

f (x) = \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{10 s + 3}{s ^{2} + 4 s + 29}

x \to 0 lim ((2 x - 8)^{\frac{1}{3}})

\frac{d}{d x} (csc (π x))

y d x - 2 (x + y) d y = 0

\frac{( 2 s - 3 )}{( s ^{2} - 3 s + 2 )}

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