do you mean
y = x ln(ln x)?
if so, then
y' = ln(lnx)) + x * 1/lnx * 1/x
= ln(lnx) + 1/lnx
y = log_x(7) = ln7/lnx
if u = lnx, then
y = ln7 u^-1
y' = -ln7 u^-2 u'
...
since cosh^2 = 1+sinh^2,
sinh^2-cosh^2 = -1
so the derivative is zero
Find the derivative of:
1. Xlog base e (log base e X)
2. log base x 7
3. (sinh x)^2 - (cosh x)^2
1 answer