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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\frac{d}{d 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{d}{d x} (\frac{1}{4 x ^{3} + 5 x ^{2} - 7 x + 8})

\int x (x^{2} + 3)^{7} d x

\frac{d}{d x} (x^{2} e n (x))

y = tan (3 x)

\int \frac{1}{( x - 7 ) ^{2}} - 1 d 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)