How do you differentiate #((e^x)+x)^(1/x)#?
1 Answer
Feb 16, 2017
# d/dx (e^x+x)^(1/x)= (e^x+x)^(1/x){ (e^x+1)/(x(e^x+x)) -(ln(e^x+x))/x^2 } #
Explanation:
The best approach here is to use logarithmic differentiation to remove the variable exponent, as follows:
Let
# => ln \ y = ln \ {(e^x+x)^(1/x)} #
# " "= 1/x \ ln(e^x+x)#
We can now differentiate the LHS implicitly, and the RHS using the product rule to get:
# \ \ \ \ \ 1/y \ dy/dx = (1/x)(d/dx ln(e^x+x)) +(d/dx 1/x)(ln(e^x+x)) #
# :. 1/y \ dy/dx = (1/x)( 1/(e^x+x) * (e^x+1))) +(-1/x^2)(ln(e^x+x)) #
# :. 1/y \ dy/dx = (e^x+1)/(x(e^x+x)) -(ln(e^x+x))/x^2 #
# :. dy/dx = (e^x+x)^(1/x){ (e^x+1)/(x(e^x+x)) -(ln(e^x+x))/x^2 } #