How do you differentiate #e^(1/(2x))#?

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Answer 1 · 2017-06-05T21:43:15

#-1/(2x^2)(e^(1/(2x)))#

Using the chain rule,

#d/dx(e^f(g))=e^f(x)xxf'(x)#

So in this case, #f(x)=(2x)^(-1)#.

The derivative of this function is

#f'(x)=-(2x)^-2xx2=-2/(2x)^2=-2/(4x^2)=-1/(2x^2)#

Therefore,

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