Question

I need help with 4B please. Thank you!

Answer 1

The equation to the tangent line to $y = e^{(x^2)}$ 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 $(x_0, y_0)$, the equation of the tangent line to this function is given by:

$y - y_0 = f^{\prime}(x_0, y_0)(x - x_0)$

In which f' is the derivative.

In this problem, the function is:

$f(x) = e^{(x^2)}$

Applying the chain rule, the derivative is:

$f^{\prime}(x) = 2xe^{(x^2)}$

At x = 1, the derivative is:

$f^{\prime}(1) = 2(1)e^{((1)^2)} = 2e$

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 brainly.com/question/8174665

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