Differentiate $\log_{10}x$ My attempt:

$\eqalign{
  & \log_{10}x = {{\ln x} \over {\ln 10}}  \cr 
  & u = \ln x  \cr 
  & v = \ln 10  \cr 
  & {{du} \over {dx}} = {1 \over x}  \cr 
  & {{dv} \over {dx}} = 0  \cr 
  & {v^2} = {(\ln10)^2}  \cr 
  & {{dy} \over {dx}} = {{\left( {{{\ln 10} \over x}} \right)} \over {2\ln 10}} = {{\ln10} \over x} \times {1 \over {2\ln 10}} = {1 \over {2x}} \cr} $


The right answer is: ${{dy} \over {dx}} = {1 \over {x\ln 10}}$ , where did I go wrong? 
Thanks!

$${\rm{lo}}{{\rm{g}}_{10}}x = {{\ln x} \over {\ln 10}} = \dfrac{1}{\ln(10)}\ln x$$
No need for the chain rule, in fact, that would lead you to your mistakes, since $\dfrac 1 {\ln(10)}$ is a constant.
So we differentiate only the term that's a function of $x$: $$\dfrac{1}{\ln(10)}\frac d{dx}(\ln x)= \dfrac 1{x\ln(10)}$$
