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Find an equation of the tangent line to the curve y = 3e⁽ˣ²⁾ at the point (1, 3e²).

User Argote
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1 Answer

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Final answer:

To find the equation of the tangent line to the curve at the point (1, 3e²), calculate the derivative to get the slope at that point and then use the point-slope form to write the equation. The slope is 6e and the equation is y - 3e² = 6e(x - 1).

Step-by-step explanation:

To find the tangent line to the curve y = 3ex² at the point (1, 3e²), we first need to calculate the derivative of y with respect to x to get the slope of the tangent line at that point. The derivative, denoted as y', is given by using the chain rule:

y = 3ex²

y' = d/dx(3ex²) = 3ex² × 2x = 6xex²

Plugging in x = 1, we find:

y'(1) = 6×1e1² = 6e

This is the slope of the tangent line at x = 1. Now we use the point-slope form of a line to write the equation of the tangent line:

y - y1 = m(x - x1)

Where (x1, y1) is the point (1, 3e²) and m is the slope 6e.

y - 3e² = 6e(x - 1)

That is the equation of the tangent line at the given point.

User Psagers
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