Step-by-step explanation:
are you sure we need the derivative of x×log(x) ?
and not of x×ln(x) ?
because it would be easier with the natural logarithm.
well, I have to take you literally.
in any case, we need to take this as a combination of simpler functions :
f(x) = x
g(x) = log(x)
h(x) = x×log(x) = (f×g)(x)
h' = f'×g + f×g' = 1×log(x) + x×(log(x))'
now, let's convert all the general logs (to the base of 10, right ?) to ln.
remember,
loga(b) = logc(b)/logc(a)
so,
log(x) = ln(x)/ln(10)
h' = ln(x)/ln(10) + x×(ln(x)/ln(10))' =
= ln(x)/ln(10) + (x/ln(10))×(ln(x))' =
= ln(x)/ln(10) + (x/ln(10))×(1/x) =
= ln(x)/ln(10) + 1/ln(10) =
= log(x) + ln(e)/ln(10) = log(x) + log(e)
FYI
if you needed the derivative of x×ln(x) after all, with the same approach we have f = x, g = ln(x) and get
h' = x×(ln(x))' + x'×ln(x) = x×(1/x) + 1×ln(x) = 1 + ln(x)
which is simply the same, if we replace log by ln in our first solution :
ln(x) + ln(e) = ln(x) + 1
