Question

Calculate y' using chain rule. y = (e ^ (1 / x))/(x ^ 2)

Answer 1

The answer is
`y^(') =e^{(1)/(x) } (2x-1)`

Explanation:

First of all, we have product of 2 functions
`e^(1/x)` and
`x^(2)`

`y^(') =u(dv)/(dx)+v(du)/(dx)`

Let
`u=e^(1/x)` and
`v=x^(2)`

using chain rule for
`u`,
`(du)/(dx)=-(1)/(x^2)`

and
`(dv)/(dx)=2x`

Therefore,

`y^(') =u(dv)/(dx)+v(du)/(dx)=e^(1/x)(2x)+x^(2) (-(1)/(x^2))=e^(1/x)(2x-1)`