How do you differentiate # f(x)=sqrt(ln(xe^x))# using the chain rule.?

1 Answer
Dec 13, 2015

Answer:

#(1+x)/(2xsqrt(ln(xe^x)))#

Explanation:

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)))#