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

1 Answer

Jan 2, 2016

We can rename #u=t^(-1/2)# and proceed, following chain rule statement: #(dy)/(dx)=(dy)/(du)(du)/(dx)#

Explanation:

now, we have #f(t)=1/u=u^-1#

#(df(t))/(dt)=-1/u^2(-1/(2t^(3/2)))=1/(((t^(-1/2))^2)(2t^(3/2)))=1/(t^(-1)*2t^(3/2))#

#(df(t))/(dt)=1/(2t^(-1+3/2))=1/(2t^(1/2))#

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