The chain rule states that if you have two nested functions, where #f(x) = g(h(x))#, the derivative is;
# f'(x) = g'(h(x)) * h'(x)#
When you have functions with many nested radicals, its often easiest to go through each function step by step.
#d/dx sqrt(ln(xe^x))#
The first function is the square root function. We can rewrite a square root as;
#sqrt(f(x)) = f(x)^(1/2)#
Using the power rule;
#d/dx f(x)^(1/2) = 1/2 f(x)^(-1/2) *f'(x)#
#= 1/(2f(x)^(1/2)) *f'(x)#
#= 1/(2sqrt(f(x))) *f'(x)#
Let #f(x) = ln(xe^x)#.
#d/dx sqrt(ln(xe^x)) = 1/(2sqrt(ln(xe^x))) d/dx ln(xe^x)#
Next we look at the #ln#. The derivative of #ln(f(x))# is #1"/"f(x) * f'(x)#.
#1/(2sqrt(ln(xe^x))) d/dx ln(xe^x) = 1/(2sqrt(ln(xe^x))) 1/(xe^x) d/dx xe^x#
The last part of the derivative can be solved using the product rule.
#d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)#
Remember that #d/dx e^x = e^x#.
#1/(2sqrt(ln(xe^x))) 1/(xe^x) d/dx xe^x = 1/(2sqrt(ln(xe^x))) 1/(xe^x) (e^x + xe^x)#
We can simplify by canceling out all of the #e^x# terms
#1/(2sqrt(ln(xe^x))) 1/(x color(red)cancel(color(black)(e^x))) (color(red)cancel(color(black)(e^x)) + xcolor(red)cancel(color(black)(e^x))) = (1+x)/(2xsqrt(ln(xe^x)))#
