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

The derivative of 8(x-1/x)^2 is 8(2x-2/(x^3))

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

The first derivative of 8(x-1/x)^2 is 8(2x-2/(x^3))

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derivative of

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (2) = 0

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

= 8 (2 x - 0 - \frac{2}{x ^{3}})

Simplify

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

\frac{d ^{2} y}{d x ^{2}} y + 81 y = 0

y^{^{'}} = k x

x \to \infty lim (4 x + 7)

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

t \to 0 lim (\frac{5}{t} - \frac{5}{t ^{2} + t})

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