Question

Tangent to the curve y=xe^x^2 passes through (1,e) then it will also pass through

Answer 1

Answer:

Step-by-step explanation:

y = x^e^x^2

Take differentitation of it.

y'=dy/dx= e^x^2 + x(2x)e^x^2  ( By product Rule)

Now put x=1 as it is given.

We get y'= 3e.

Equation of Tangent becomes

(y-e)=3e(x-1).

Now you can check according to your options which point will pass into it.

Thanks