#y=log_5(arctan(x^x))#.
The right hand side of this expression is fairly horrible. It would be better to rearrange this and use implicit differentiation than attempt to differentiate directly.
#5^y = arctan(x^x)#.
A common trick to help differentiate functions involving exponents is to write the base as #e# and modify the power with a logarithm to give the desired exponent. This then makes differentiating easier. Notice that #x^x=e^{xln(x)}# and #5^y=e^{yln(5)}#. Then,
#e^{yln(5)}=arctan(e^{xln(x)})#.
Differentiate implicitly. Note that #"d"/("d"x) arctan(x) = 1/(x^2+1)#.
#ln(5)e^{yln(5)}("d"y)/("d"x)=1/(e^{2xln(x)}+1) "d"/("d"x) (e^{xln(x)})#.
As #"d"/("d"x) e^{xln(x)} = (1+ln(x))e^{xln(x)}#, we see that,
#("d"y)/("d"x) ln(5) e^{yln(5)} =1/(e^{2xln(x)}+1) (1+ln(x))e^{xln(x)}#.
We can now rewrite the exponents as what they were originally and rearrange for #("d"y)/("d"x)#.
#("d"y)/("d"x) = ((1+ln(x))x^x)/((x^{2x}+1)ln(5)5^y)#.
We know from the definition that #5^y=arctan(x^x)#.
We conclude,
#("d"y)/("d"x) = ((1+ln(x))x^x)/(ln(5)(x^{2x}+1)arctan(x^x))#.
