Question #5e314

Well, I would apply the product rule. Normally one might suggest the quotient rule, but I'm just used to the product rule, and it's also a nice alternative if the quotient rule appears to be cumbersome. I did it both ways though.

Define:

#f(x) = g(x)cdoth(x)#

Then, the product rule says that the derivative is given by:

#(df)/(dx) = g(x)(dh)/(dx) + h(x)(dg)/(dx)#

So, with #g(x) = 1 - lnx# and #h(x) = 1/(1 + lnx)#:

#color(blue)(d/(dx)[(1 - lnx)/(1 + lnx)])#

#= [overbrace((1 - lnx))^(g(x)) cdot overbrace(-1/(1 + lnx)^2 cdot 1/x)^(dh//dx)] + [overbrace(1/(1 + lnx))^(h(x))cdot overbrace(-1/x)^(dg//dx)]#

#= -(1 - lnx)/(x(1 + lnx)^2) - 1/(x(1 + lnx))#

Now we achieve common denominators by multiplying the right-hand fraction by #(1 + lnx)/(1 + lnx)#:

#=> -(1 - lnx)/(x(1 + lnx)^2) - (1 + lnx)/(x(1 + lnx)^2)#

And be careful with this step... many a negative has tripped the masters.

#=> -(1 - cancel(lnx) + 1 + cancel(lnx))/(x(1 + lnx)^2)#

#= color(blue)(-(2)/(x(1 + lnx)^2))#

Alternatively, one could use the quotient rule.

Define:

#f(x) = (g(x))/(h(x))#

Then, with #g(x) = 1 - lnx# and #h(x) = 1 + lnx#:

#color(blue)((df)/(dx)) = (h(x)(dg)/(dx) - g(x)(dh)/(dx))/(h(x))^2#

#= ((1 + lnx)(-1/x) - (1 - lnx)(1/x))/((1 + lnx)^2)#

#= -1/x((1 + cancel(lnx) + 1 - cancel(lnx))/((1 + lnx)^2))#

#= color(blue)(-2/(x(1 + lnx)^2))#