Question

How do you find the equation of the tangent line to the graph `f(x)=ln(e^(x^2))` through point (-2,4)?

Answer 1

`4x+y+4=0`

Explanation:

As (e) and (ln) are inverse functions,

`y = f(x)=ln(e^(x^2))=x^2`

So, y' = 2x = -4, at P(-2. 4).

The equation of the tangent at P is

`y-4=-4(x-(-2))`. Simplifying,

`4x+y+4=0`.