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

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Answer 1 · 2016-02-29T18:52:36

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

#\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}#