Find an equation of the tangent line to the curve at the given point.
y = 4e^xcos(x), (0, 4)
dy/dx = 4 [ e^x (-sin x) + cos x (e^x) ]
at x = 0
dy/dx = m = 4 [ 0 + 1 ] = 4
so
y = 4 x + b
at x = 0, y = 4
4 = 4(0) + b so b = 4
y = 4 x + 4
Oh boy, we got ourselves a curve! Let's put on our clown shoes and find that tangent line, shall we?
To find the equation of the tangent line at the point (0, 4), we need two things: the derivative of the curve and the point itself. So let's start by taking the derivative of y = 4e^xcos(x).
The derivative of y with respect to x is given by the chain rule. You might say it's a "derivative circus act!" We have to differentiate 4e^x with respect to x, which is just 4e^x. Then, we need to multiply it by the derivative of cos(x), which is -sin(x). So, the derivative of y is:
dy/dx = 4e^x(-sin(x)).
Now that we have the derivative, we can use the point (0, 4) to find the slope of the tangent line. So let's plug in x = 0 into the derivative:
dy/dx = 4e^0(-sin(0)) = 0.
Oh no, we got ourselves a flat line! The slope is 0. It's like a clown without any tricks up its sleeve!
Now that we know the slope of the tangent line is 0, we can use the point-slope form of the equation of a line:
(y - y1) = m(x - x1),
where m is the slope and (x1, y1) is the given point. Plugging in the values, we get:
(y - 4) = 0(x - 0) = 0.
Alright, looks like we have a straight line equation here! The equation of the tangent line to the curve y = 4e^xcos(x) at the point (0, 4) is simply y = 4.
And there you have it, my comedic friend! The tangent line is as straight as a clown's smile. If you have any more mathematical wonders to uncover, feel free to ask!
To find the equation of the tangent line to the curve y = 4e^xcos(x) at the given point (0, 4), we need to find the slope of the tangent line and then use the point-slope form of a line.
1. Find the derivative of the function y = 4e^xcos(x) with respect to x.
Differentiating y = 4e^xcos(x) using the product rule:
dy/dx = (4 * e^x * cos(x)) + (4 * e^x * (-sin(x)))
= 4e^x * (cos(x) - sin(x))
2. Plug in the x-coordinate of the given point (0, 4) to find the slope of the tangent line.
Slope (m) = dy/dx evaluated at x = 0
= 4e^0 * (cos(0) - sin(0))
= 4(1)(1 - 0)
= 4
3. Now we have the slope (m = 4) and the point (0, 4).
Using the point-slope form: y - y1 = m(x - x1)
where (x1, y1) is the given point.
Plugging in the values, we get:
y - 4 = 4(x - 0)
y - 4 = 4x
4. Simplify the equation:
y = 4x + 4
Therefore, the equation of the tangent line to the curve y = 4e^xcos(x) at the point (0, 4) is y = 4x + 4.
To find the equation of the tangent line to a curve at a given point, we need to find the slope of the tangent line and use the point-slope form equation.
1. First, find the derivative of the given function using the product rule and chain rule. In this case, the function is y = 4e^x * cos(x):
dy/dx = (d/dx)(4e^x * cos(x))
= 4(e^x * cos(x))' + 4(cos(x) * e^x)'
= 4(e^x * cos(x) - sin(x) * e^x)
2. Now evaluate the derivative at the given x-coordinate to find the slope of the tangent line. The point is (0, 4), so x = 0:
slope = dy/dx | (x=0)
= 4(e^0 * cos(0) - sin(0) * e^0)
= 4(cos(0) - sin(0))
= 4(1 - 0)
= 4
Therefore, the slope of the tangent line is 4.
3. Now we can use the point-slope form equation to find the equation of the tangent line. The point given is (0, 4) and the slope is 4:
y - y1 = m(x - x1)
Plugging in the values: y - 4 = 4(x - 0)
Simplifying, we get: y - 4 = 4x
Rearranging, the equation of the tangent line is: y = 4x + 4.