What is the derivative of (8x^2+9)(x^2-1/x) ?

The derivative of (8x^2+9)(x^2-1/x) is 32x^3+18x+9/(x^2)-8

What is the first derivative of (8x^2+9)(x^2-1/x) ?

The first derivative of (8x^2+9)(x^2-1/x) is 32x^3+18x+9/(x^2)-8

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derivative of

(8 x^{2} + 9) (x^{2} - \frac{1}{x})

32 x^{3} + 18 x + \frac{9}{x ^{2}} - 8

Solution steps

\frac{d}{d x} ((8 x^{2} + 9) (x^{2} - \frac{1}{x}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 8 x^{2} + 9, g = x^{2} - \frac{1}{x}

= \frac{d}{d x} (8 x^{2} + 9) (x^{2} - \frac{1}{x}) + \frac{d}{d x} (x^{2} - \frac{1}{x}) (8 x^{2} + 9)

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

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

= 16 x (x^{2} - \frac{1}{x}) + (2 x + \frac{1}{x ^{2}}) (8 x^{2} + 9)

Simplify

16 x (x^{2} - \frac{1}{x}) + (2 x + \frac{1}{x ^{2}}) (8 x^{2} + 9) : 32 x^{3} + 18 x + \frac{9}{x ^{2}} - 8

= 32 x^{3} + 18 x + \frac{9}{x ^{2}} - 8

6 x^{2} - 4 x + 5

\int \frac{( x ^{4} - 5 x ^{2} + 3 x - 4 )}{x} d x

\int_{1}^{e} (\frac{x ^{2} - 1}{x}) d x

\frac{d}{d x} (ln (\frac{x + y}{x - y}))

\frac{\partial}{\partial s} (sin (s^{2} - t^{2}))

What is the derivative of (8x^2+9)(x^2-1/x) ?
The derivative of (8x^2+9)(x^2-1/x) is 32x^3+18x+9/(x^2)-8
What is the first derivative of (8x^2+9)(x^2-1/x) ?
The first derivative of (8x^2+9)(x^2-1/x) is 32x^3+18x+9/(x^2)-8