So, I've got such a function: y = x^(log base5 (x)), which I have to differentiate in terms of ln;
I started by solving the equation y = x^(lnx/ln5) = x^(lnx - ln5) (wonder if here is my mistake)
after that ln y = ln x^(lnx - ln5)=> 1/y*y' = (lnx - ln5)'(ln x) + (ln x)'(lnx - ln5) => (1/x - 1/5)(ln x) + (1/x)(lnx - ln5) =>
y' = x^(lnx - ln5)[(1/x - 1/5)(ln x) + ((lnx - ln5)/x)], that's what I got, but putting it in webwork it consideres it as incorrect,
tell me please were is my mistake,
Thanks,
so, I assume the right answer would be
y'/y = [((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + (lnx/ln5)*(1/x) =>
y' = (lnx/ln5)*ln(x) {[((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + (lnx/ln5)*(1/x)}
y' = (lnx/ln5)*ln(x){[((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + ((x lnx/ln5)}, there is a way to contract it a little bit or its better to leave it like this?
oh, let me just continue, y' = (lnx/ln5)*ln(x){[((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + ((x lnx/ln5)}, =>
y' = (lnx/ln5)*ln(x){[ (ln 5/x) - (ln x /5)/(ln5)^(2)]*ln x + ((x lnx/ln5)}, =>
y' = (lnx/ln5)*ln(x) [((lnx/((ln5)^(2))*(ln5/x)-(lnx/5)) + ((x lnx/ln5)]
does that makes sense?