Question

u=x^y +y^x differentiate wrt x

Answer 1

Let s= x^y

t= y^x

u= s+t

u'= s'+t'

s=x^y

log s = y logx

1/s×s' = y'× logx + y/x

s' = s[y' × logx + y/x]

t=y^x

log t = x logy

1/t × t' = logy + x/y × y'

t' = t[logy + xy'/y]

Therefore,

u' = s[y' × logx + y/x] + t[logy + xy'/y]

u' = x^y [y' × logx + y/x] + y^x [logy + xy'/y]