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

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

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

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

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derivative of

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

\frac{- 16 x ^{3} + 9}{( 8 x ^{3} - 2 x + 9 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{- 16 x ^{3} + 9}{( 8 x ^{3} - 2 x + 9 ) ^{2}}

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