How do you differentiate #f(x)=sqrt(1/x^2)# using the chain rule?

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Answer 1 · 2017-03-05T22:18:19

#d/dx sqrt(1/x^2) = - absx/x^3#

You can name:

#y(x) = 1/x^2 = x^-2#

so that:

#(df)/dx = (df)/(dy)dy/dx = d/dy(sqrt(y))d/dx(x^-2) = 1/(2sqrt(y))(-2x^-3) = 1/(2sqrt(1/x^2))(-2x^-3) = -sqrt(x^2)/x^3 = - abs(x)/x^3#

You can also note that:

#f(x) = sqrt(1/x^2) = 1/absx#

so that:

#{((df)/dx = d/dx (1/x) = -1/x^2" for " x > 0), ((df)/dx = d/dx (-1/x) = 1/x^2" for " x < 0):}#

which is clearly the same.