What is the derivative of (2x^2-7)/(8x^3+1) ?

The derivative of (2x^2-7)/(8x^3+1) is (-16x^4+168x^2+4x)/((8x^3+1)^2)

What is the first derivative of (2x^2-7)/(8x^3+1) ?

The first derivative of (2x^2-7)/(8x^3+1) is (-16x^4+168x^2+4x)/((8x^3+1)^2)

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derivative of

\frac{2 x ^{2} - 7}{8 x ^{3} + 1}

\frac{- 16 x ^{4} + 168 x ^{2} + 4 x}{( 8 x ^{3} + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{2} - 7) = 4 x

\frac{d}{d x} (8 x^{3} + 1) = 24 x^{2}

= \frac{4 x ( 8 x ^{3} + 1 ) - 24 x ^{2} ( 2 x ^{2} - 7 )}{( 8 x ^{3} + 1 ) ^{2}}

Expand

4 x (8 x^{3} + 1) - 24 x^{2} (2 x^{2} - 7) : - 16 x^{4} + 168 x^{2} + 4 x

= \frac{- 16 x ^{4} + 168 x ^{2} + 4 x}{( 8 x ^{3} + 1 ) ^{2}}

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

n = 1 \sum \infty \frac{- 1 ^{n + 1}}{n}

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

\int_{0}^{1} \frac{6}{1 + t ^{2}} d t

\frac{d}{d x} (\frac{cos ( x )}{x} + \frac{x}{cos ( x )})

What is the derivative of (2x^2-7)/(8x^3+1) ?
The derivative of (2x^2-7)/(8x^3+1) is (-16x^4+168x^2+4x)/((8x^3+1)^2)
What is the first derivative of (2x^2-7)/(8x^3+1) ?
The first derivative of (2x^2-7)/(8x^3+1) is (-16x^4+168x^2+4x)/((8x^3+1)^2)