Question

How do you differentiate `f(x)=e^sqrt(1/x)` using the chain rule.?

Answer 1

Please follow the instructions below...

Explanation:

`y=e^u`

Use part of the chain rule...

`:. (dy)/(du)=e^u=e^(sqrt(1/x))`

`u=sqrt(1/x)`

`u=(1/x)^(1/2)`

`u^2=1/x`

`ln(u^2)=ln(1/x)`

`2ln(u)=ln1-lnx`

`2lnu=-lnx`

Now use implicit differentiation...

`2/u*(du)/(dx)=-1/x`

`u/2*2/u*(du)/(dx)=-1/x*u/2`

`(du)/(dx)=-1/(2x)*sqrt(1/x)`

Now use the chain rule...

`(dy)/(du)*(du)/(dx)=-1/(2x)*e^(sqrt(1/x))*sqrt(1/x)=(dy)/(dx)`

Now, you can simplify the final result...

`-1/(2x)*e^(sqrt(1/x))*sqrt(1/x)`

`=-1/(2x)*e^(sqrt(1/x))*sqrt(1)/sqrt(x)*sqrt(x)/sqrt(x)`

`=-1/(2x)*e^(sqrt(1/x))*sqrt(x)/x`

`=-1/(2x^2)*e^(sqrt(1/x))*sqrt(x)`

Still looks quite ugly, but this is the result you're looking for. It is what it is.

Answer 2

`e^(sqrt(1/x))/(2x^(3/2)`

Explanation:

differentiate using the `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|))).... (A)`

let `u=sqrt(1/x)=(1/x)^(1/2)=1/x^(1/2)=x^(-1/2)`

differentiate using the `color(blue)"power rule"`

`rArr(du)/(dx)=-1/2x^(-3/2)`

Now `y=f(x)=e^urArr(dy)/(du)=e^u`

substitute results for `(dy)/(du)" and " (du)/(dx)` into (A) changing u back to x.

#rArrdy/dx=e^u.-1/2x^(-3/2)=-1/2x^(-3/2)e^(sqrt(1/x))#

`rArrdy/dx=(e^(sqrt(1/x)))/(2x^(3/2)`