Question

Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the given point.

y = e^2x^2, (1, e^2)

Answer 1

`y = 4e^2x - 3e^2`

Explanation:

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

lets calculate the derivate of f using the chain rule:

`f'(x) = e^(2x^2)* (2x^2)' = e^(2x^2)*4x = 4e^(2x^2)x`

we have that f'(1) = 4e^2, hence the equation is

`y = f(1) + f'(1) (x-1) = e^2 + 4e^2(x-1)`

or, equivalently,

`y = 4e^2x - 3e^2`