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

2 Answers
Oct 7, 2016

Answer:

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.

Oct 7, 2016

Answer:

#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)#