The equation to the tangent line to [tex]y = e^{(x^2)}[/tex] at point (1,0) is given by:
y = 2e(x - 1).
What is the equation to the tangent line of a function of a point?
Supposing that we have a function f(x) at a point [tex](x_0, y_0)[/tex], the equation of the tangent line to this function is given by:
[tex]y - y_0 = f^{\prime}(x_0, y_0)(x - x_0)[/tex]
In which f' is the derivative.
In this problem, the function is:
[tex]f(x) = e^{(x^2)}[/tex]
Applying the chain rule, the derivative is:
[tex]f^{\prime}(x) = 2xe^{(x^2)}[/tex]
At x = 1, the derivative is:
[tex]f^{\prime}(1) = 2(1)e^{((1)^2)} = 2e[/tex]
Hence the tangent line at point (1,0) is given by:
y = 2e(x - 1).
More can be learned about the equation of a tangent line at https://brainly.com/question/8174665
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