Differentiate `x^x^(2-3)+(x-3)^x^2` with respect to `x` :

Explanation

Let `y=x^(x^(2)-3)+(x-3)^(x^(2))`
Let `u=x^(x^(2)-3)" and "v=(x-3)^(x^(2))`
` :. y=u+v`
`implies(dy)/(dx)=(du)/(dx)+(dv)/(dx) " " `...(1)
Now, `u=x^(x^(2))-3`
`implieslogu=logx^(x^(2)-3)=(x^(2)-3)logx`
`implies(1)/(u)(du)/(dx)=(x^(2)-3)(d)/(dx)logx+logx*(d)/(dx)(x^(2)-3)`
`implies(du)/(dx)=u[(x^(2)-3)/(x)+2xlogx]`
`=x^(x^(2)-3)*[(x^(2)-3)/(x)+2xlogx]`
and `v=(x-3)^(x^(2))`
`implieslogv=log(x-3)^(x^(2))=x^(2)log(x-3)`
`implies(1)/(v)(dv)/(dx)=x^(2)*(d)/(dx)log(x-3)+log(x-3)(d)/(dx)x^(2)`
`implies(dv)/(dx)=v[(x^(2))/(x-3)+2x log(x-3)]`
`(x-3)^(x^(2))[(x^(2))/(x-3)+2x log(x-3)]`
` :.`From equation (1)
`(dy)/(dx)=x^(x^(2)-3)[(x^(2)-3)/(x)+2xlogx]+(x-3)^(x^(2))[(x^(2))/(x-3)+2xlog(x-3)]`