I don't know if you are using #log# for #log_10# or for the natural log.
I would not use the chain rule. I define #lnx# as a definite integral, then define #e^x# as the inverse function of #lnx#. Then #10^x = e^(xln10)#
If we define #y=logx# iff and only if #10^y=x#
It is not difficult to show that #logx=lnx/ln10# so that #d/(dx)( logx)=1/ln10*1/x# But I did not use the chain rule there.
Using #y=logx# iff and only if #10^y=x# We might differentiate implicitly, which is a way of using the chain rule:
#d/(dx)(10^y)=d/(dx)(x)#
#10^y ln10 (dy)/(dx) = 1#
So, #(dy)/(dx) = 1/(10^y ln10) = 1/xln10#
If you're using #logx# for the inverse of #e^x#, then you could again use implicit differentiation, which is really the chain rule, to find the derivative.
