Question

Find the equation of the tangent line to the curve at the given point?

ln(xe^x^2) at point (1,1)

Answer 1

`y=3(x+1)+1`

Explanation:

Step 1 : Differentiate
`y'= (1/(xe^{x^2}) * (e^{x^2} + xe^{x^2} *2x)`

Step 2: To find the slope of the tangent line, we need to evaluate y' at our given point:

`y'(1) = (1/e) * (e+e*2)`
`y'(1) = 3`

Step 3: using the slope we just found and the point given, we can find the equation of the tangent line at `(1,1)`:
`y=3(x+1)+1`