Find the equation of a normal to the curve `y=x(log)_e x` which is parallel to the line `2x-2y+3=0` .

Explanation

Slope of the given line is 1

Let `(x_{1}, y_{1})` be the point where the tangent is drawn to the curve.

Since, the point lies on the curve.

Hence,`y_{1}=x_{1} log _{e} x_{1} ldots ldots ldots` (1)

Now, `y=x log _{e} x`

`Rightarrow frac{d y}{d x}=x times frac{1}{x}+log _{e} x(1)=1+log _{e} x`

Slope of tangent `=1+log _{e} x_{1}`

Slope of normal `=frac{-1}{text { Slope of tangent }}=frac{-1}{1+log _{e} x_{1}}`

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