How do you use the chain rule to differentiate #y=(x-1/x)^(3/2)#?

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Answer 1 · 2017-10-24T11:29:55

#color(blue)( 3/2sqrt(x-1/x)(1+1/x^2))#

Write #(x-1/x)^(3/2)# as #(x-x^-1)^(3/2)#

Let #(x-x^-1) = u#

Then #y= u^(3/2)#

The chain rule states that:

#dy/dx= dy/(du) * (du)/dx#

So:

#3/2(u)^(1/2)*(1-(-x^-2)#

#3/2(x-x^-1)^(1/2)*(1+x^-2)=color(blue)( 3/2sqrt(x-1/x)(1+1/x^2))#