Differentiate w.r.t. x the function `x^(x^(2)-3)+(x-3)^(x^2)` for `x > 3`.

Explanation

Let `y=x^(x^(2)-3)+(x-3)^(x^2)`

And let `u=x^(x^(2)-3)`, `v= (x-3)^(x^2)`

Now,

`y=u+v`

Differentiating both sides w.r.t. x

`(dy)/(dx)=(d(u+v))/dx`

`(dy)/(dx)=(du)/dx+(dv)/dx`


Calculating `(du)/dx`

`u=x^(x^(2)-3)`

Taking log on both sides

`log u=logx^(x^2-3)`

`log u=(x^2-3)logx`

Differentiating both sides w.r.t. x

`(d(log u))/dx=(d((x^2-3)logx))/dx`

`(d(log u))/dx.(du)/(du)=(d((x^2-3)logx))/dx`

`(d(log u))/(du).(du)/(dx)=(d((x^2-3)logx))/dx`

`1/u.(du)/(dx)=(d((x^2-3)logx))/dx`

Using product rule

`1/u.(du)/(dx)=(d(x^2-3))/dx.logx+(d(logx))/dx.(x^2-3)`

`1/u.(du)/(dx)=(2x-0)logx+1/x xx(x^2-3)`

`1/u.(du)/(dx)=2xlogx+(x^2-3)/x `

`(du)/(dx)=u(2xlogx+(x^2-3)/x )`

`(du)/(dx)=x^(x^(2)-3)(2xlogx+(x^2-3)/x )`


Calculating `(dv)/dx`

`v= (x-3)^(x^2)`

Taking log on both sides

`logv=log(x-3)^(x^2)`

`logv=x^2.log(x-3)`

Differentiating both sides w.r.t. x

`(d(logv))/dx=(d(x^2.log(x-3)))/dx`

`(d(logv))/dx.(dv)/(dv)=(d(x^2.log(x-3)))/dx`

`(d(logv))/(dv).(dv)/(dx)=(d(x^2.log(x-3)))/dx`

`1/v.(dv)/(dx)=(d(x^2.log(x-3)))/dx`

Using product rule

`1/v.(dv)/(dx)=(d(x^2))/dx.log(x-3)+(d(log(x-3)))/dx.x^2`

`1/v.(dv)/(dx)=2x.log(x-3)+1/(x-3).(d(x-3))/dx.x^2`

`1/v.(dv)/(dx)=2x.log(x-3)+1/(x-3).x^2`

`(dv)/(dx)=v(2x.log(x-3)+x^2/(x-3))`

`(dv)/(dx)=(x-3)^(x^2)(2x.log(x-3)+x^2/(x-3))`


Now,

`(dy)/(dx)=(du)/dx+(dv)/dx`

`(dy)/(dx)=x^(x^(2)-3)(2xlogx+(x^2-3)/x )+(x-3)^(x^2)(2x.log(x-3)+x^2/(x-3))`