How do you find the derivative of #y=e^(1/x)#?

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Answer 1 · 2017-10-15T13:32:29

#dy/dx =-e^(1/x)/x^2#

Use the chain rule by posing #u = 1/x#, so that #y=e^u#:

#dy/dx = dy/(du) (du)/dx = d/(du) e^u xx d/dx 1/x = e^u xx -1/x^2 =-e^(1/x)/x^2#

Answer 2 · 2017-10-15T13:39:27

Let's see.

Let the equation #y=e^(1/x)# be a function of #x# such that #rarr#

#y=f(x)=e^(1/x)#

Now, differentiating the equation w.r.t #x# we get #rarr#

#dy/dx=d/dx(e^(1/x))#

#:.dy/dx=e^(1/x)cdotd/dx(1/x)# #larr# Chain Rule.

#:.dy/dx=e^(1/x)cdot(-1/x^2)#

#:.color(red)(dy/dx=-e^(1/x)/x^2)#. (Answer).

Hope it Helps:)