Question

For `f(x)=e^(xlnx)` what is the equation of the tangent line at `x=1`?

Answer 1

`y=x`

Explanation:

`f(1)=e^(1ln1)=1`

`f'(x)=e^(xlnx)*(x*1/x+lnx)=(1+lnx)*e^(xlnx)`

`f'(1)=1`

Therefore tangent line is straight line so of form `y=mx+c`, with gradient `m=1`.
Substituting point `(1,1)` into equation of tangent yields
`1=(1)(1)+c=>c=0`.

Hence equation of required tangent is `y=x`.