Question

find d/dx [log_x_e], where x is the base, answer should be "-1/x((lnx)^2)" but I need to know the steps and process

Answer 1

log_x_e = \[\log_{e} e \over \log_{e} x\] = \[\ln e \over \ln x\] but ln e = 1 therfore \[1 \over \ln x\] Now what remains is to find the derivative:
\[{{d \over dx} ({1 \over \ln x})} = {{\ln x * {d(1) \over dx} - 1 * {{d (\ln x) } \over {dx}}} \over \ln ^2 x} = {\ln x * 0 - 1 * {1 \over x} \over \ln^2 x} = {{-1 \over x} \over \ln^2 x} = {-1 \over x * \ln ^ 2 x}\] Which is your solution :)

Answer 2

I changed base in the first step conveniently selecting the base as e \[\log_{x} e = {\log_{e} e \over \log_{e} x}\]
but \[\log_{e} a = \ln a\] therefore \[\ln e \over \ln x\]
after which I used the Quotient rule for differentiation.
I hope I explained it clearly enough :)