Question

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

Answer 1

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))`