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

Question

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

1528 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-23T13:29:40

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

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 $(1,1)$ :
$y=3(x+1)+1$