Expert: Sherry Wallin - 2/13/2009

Question
(i) Fine dy/dx,if f(x)=xlnx
(ii) If x= n/rln(1/r),find the value of r for which x will be maximum

Answer
Hi Tony~
    For i) just use the product rule for derivatives. When asked to find dy/dx it is the same thing as saying find the derivative of y with respect to x and remember y = f(x) so you really have y = xlnx and so on the left the derivative of y is y' and on the right side the derivative is, (using the product rule g(x)h'(x) + h(x)g'(x) and let g(x) = s and h(x) = ln x)
x(1/x) + (ln x)(1) = 1 + ln x; really it is as simple as that

For ii) You will need to take the derivative to find the maximum of the function. Here you have two variables, x and r and x is the dependent variable (instead of y) and r is the independent variable (instead of x). when you are dealing with x and y think of the xy coordinant system and ordered pairs being (x,y), here you have an rx coordinant system with ordered pairs (r,x). In essence instead of having y = f(x) you have x = f(r). So calculating the derivative on the left you have x' and on the right you need to use the quotient rule [I am assuming you mean n/[r(ln(1/r))] letting g(r) = n and h(r) = rln(1/r) and noting n is a constant and h(r) is again a product composed of j(r)= r and k(r)= ln(1/r). The form of the quotient rule I use is the bottom function times the derivative of the top function minus the top function times the derivative of the bottom function, all divided by the bottom function squared:
First let's just calculate the individual derivatives:
r(ln(1/r)) = r(1/(1/r) + ln(1/r)(1) = r^2+ln(1/r)
now put in the values in the quotient rule:

[r(ln(1/r)(o)-n(r^2+ln(1/r))]/[r(ln(1/r)]^2
= -n(r^2+ln(1/r))/[r(ln(1/r)]^2
remember this is x' and in calculus you know that when the derivative is 0 that is where you will get your min or max so find the solution to -n(r^2+ln(1/r))/[r(ln(1/r)]^2  = 0

Have a fruitful journey,

Math Prof