What is the derivative of $y = \frac{1}{2} (x^{2} - x^{- 2})$ $y=1/2(x^2-x^-2)$ ?

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Answer 1 · 2016-02-29T18:52:36

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

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

Taking the constant out as: $(a\cdot f)^'=a\cdot f^'$

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

Applying sum/difference rule s: $(f\pm g)^'=f^'\pm g^'$

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

We have,
$d/dx (x^2)$ = $2x$

$d/dx((x^(-2))$ = $-2x^(-3)$ = $-2/x^3$

Finally,
$=\frac{1}{2}(2x-(-\frac{2}{x^3}))$

Simplifying it,we get,

$x+\frac{1}{x^3}$

What is the derivative of y=1/2(x^2-x^-2)y=12(x2−x−2)y=1/2(x^2-x^-2)?