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

Answer:

$y = 4e^2x - 3e^2$  

Step-by-step 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,

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