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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)

Respuesta :

Answer:

[tex] y = 4e^2x - 3e^2 [/tex]  

Step-by-step explanation:

[tex] f(x) = e^{2x^2} [/tex]

lets calculate the derivate of f using the chain rule:

[tex] f'(x) = e^{2x^2}* (2x^2)' = e^{2x^2}*4x = 4e^{2x^2}x [/tex]

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

[tex] y = f(1) + f'(1) (x-1) = e^2 + 4e^2(x-1) [/tex]

or, equivalently,

[tex] y = 4e^2x - 3e^2 [/tex]  

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