Question

Find an equation of the tangent line to the curve at the given point. y = 4e^x cos(x), (0, 4)

Answer 1

The equation of the tangent line to the curve y = 4eˣ cos(x) at the point (0, 4) is y = 4x + 4.

Step-by-step explanation:

To find the equation of the tangent line to the curve y = 4eˣ cos(x) at the given point (0, 4), we need to follow these steps:

1. Determine the derivative of the function, which represents the slope of the tangent line at any point on the curve.
2. Calculate the slope at the specific point by plugging in the x-coordinate of the point into the derivative.
3. Use the point-slope form of a line with the slope from step 2 and the point provided to write the equation of the tangent line.

First, let's find the derivative of the function using the product rule since y is the product of two functions, 4eˣ and cos(x). The derivative of y with respect to x is:

y' = d/dx(4eˣ cos(x)) = 4eˣ d/dx[cos(x)] + cos(x) d/dx[4eˣ]

y' = 4eˣ (-sin(x)) + cos(x) * 4eˣ

y' = 4eˣ(-sin(x) + cos(x))

Calculating the derivative at the point (0, 4), we substitute x with 0:

y'(0) = 4e⁰(-sin(0) + cos(0)) = 4(0 + 1) = 4

The slope of the tangent line at x = 0 is 4. Next, we use the point-slope form of a line equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the curve.

The equation of the tangent line is thus:

y - 4 = 4(x - 0)

y = 4x + 4