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

Question

Isaiah Allen · Accepted Answer

# 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 } #

The best approach here is to use logarithmic differentiation to remove the variable exponent, as follows: Let # y = (e^x+x)^(1/x) # # => 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 } #

Isaiah Allen · Answer

undefined To differentiate ((e^x)+x)^(1/x), you can use the chain rule. First, take the natural logarithm of the function to simplify it. Then apply the chain rule and differentiate step by step. The result will be a bit complex, involving the original function and its derivatives. Here's the simplified process:

Let y = ((e^x) + x)^(1/x)
Take the natural logarithm of both sides: ln(y) = (1/x) * ln((e^x) + x)
Differentiate implicitly with respect to x:
(dy/dx) / y = [(1/x) * (d/dx) ln((e^x) + x)] - ln((e^x) + x) / x^2

Then solve for dy/dx. The result will involve the original function and its derivatives.