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

The derivative of y=(x+2/x)(7-2x^3) is -8x^3-8x-(14)/(x^2)+7

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

The first derivative of y=(x+2/x)(7-2x^3) is -8x^3-8x-(14)/(x^2)+7

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derivative of

y = (x + \frac{2}{x}) (7 - 2 x^{3})

- 8 x^{3} - 8 x - \frac{14}{x ^{2}} + 7

Solution steps

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

Apply the Product Rule:

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

f = x + \frac{2}{x}, g = 7 - 2 x^{3}

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

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

\frac{d}{d x} (7 - 2 x^{3}) = - 6 x^{2}

= (1 - \frac{2}{x ^{2}}) (7 - 2 x^{3}) + (- 6 x^{2}) (x + \frac{2}{x})

Simplify

(1 - \frac{2}{x ^{2}}) (7 - 2 x^{3}) + (- 6 x^{2}) (x + \frac{2}{x}) : - 8 x^{3} - 8 x - \frac{14}{x ^{2}} + 7

= - 8 x^{3} - 8 x - \frac{14}{x ^{2}} + 7

\int cos^{4} (x s i) n^{4} x

\int \frac{( 1 + x )}{( 1 - x )} d x

\frac{d y}{d t} - 0.06 y = - 24000, y (0) = 250000

f (x) = (x - 1)^{2} - 2

\frac{\partial}{\partial x} (2 y e^{7 x})

What is the derivative of y=(x+2/x)(7-2x^3) ?
The derivative of y=(x+2/x)(7-2x^3) is -8x^3-8x-(14)/(x^2)+7
What is the first derivative of y=(x+2/x)(7-2x^3) ?
The first derivative of y=(x+2/x)(7-2x^3) is -8x^3-8x-(14)/(x^2)+7