at any x the slope is y' = 6 cos6x + 12 sin6x cos6x
y'(0) = 6
so, the line through (0,0) with slope 6 is y=6x
y = sin 6x + sin2 6x, (0, 0)
y'(0) = 6
so, the line through (0,0) with slope 6 is y=6x
To find the slope of the tangent line, we need to find the derivative of the curve. So, let's differentiate y = sin 6x + sin² 6x with respect to x.
Using the chain rule, the derivative of sin 6x is 6 cos 6x, and the derivative of sin² 6x is 2 sin 6x multiplied by the derivative of 6x, which is 6.
Putting it all together, the derivative of y is:
dy/dx = 6 cos 6x + 12 sin 6x cos 6x.
Now that we have the derivative, let's substitute the given x-coordinate, which is 0, into the derivative: dy/dx at x = 0.
dy/dx at x = 0 = 6 cos (6 * 0) + 12 sin (6 * 0) cos (6 * 0)
dy/dx at x = 0 = 6 cos 0 + 12 sin 0 cos 0
dy/dx at x = 0 = 6 * 1 + 12 * 0 * 1
dy/dx at x = 0 = 6
So, the slope of the tangent line at the point (0, 0) is 6.
Now, using the point-slope form of the line, we can write the equation of the tangent line:
y - y₁ = m(x - x₁), where (x₁, y₁) is (0, 0) and m is the slope we found.
Simplifying, we get:
y - 0 = 6(x - 0)
y = 6x
Therefore, the equation of the tangent line to the curve y = sin 6x + sin² 6x at the point (0, 0) is y = 6x.
Step 1: Find the derivative of y with respect to x using the chain rule.
y = sin(6x) + sin^2(6x)
To find the derivative, we will differentiate each term separately using the chain rule.
Derivative of sin(6x): (cos(6x) * 6)
Derivative of sin^2(6x): 2(sin(6x)) * (cos(6x) * 6)
y' = (cos(6x) * 6) + 2(sin(6x)) * (cos(6x) * 6)
Step 2: Simplify the equation.
y' = 6cos(6x) + 12sin(6x) * cos(6x)
Step 3: Evaluate the derivative at x = 0 to find the slope of the tangent line at the given point (0, 0).
slope = y'(0)
= 6cos(6(0)) + 12sin(6(0)) * cos(6(0))
= 6cos(0) + 12sin(0) * cos(0)
= 6(1) + 12(0) * 1
= 6
The slope of the tangent line at the point (0, 0) is 6.
Step 4: Use the point-slope formula to find the equation of the tangent line.
Using the formula: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the curve.
Plug in the values:
m = 6
x1 = 0
y1 = 0
y - 0 = 6(x - 0)
y = 6x
Therefore, the equation of the tangent line to the curve y = sin(6x) + sin^2(6x) at the point (0, 0) is y = 6x.
Step 1: Find the derivative of the function y = sin 6x + sin^2 6x.
To do this, we need to apply the chain rule and power rule. Let's start with the first term.
The derivative of sin u, where u is a function of x, is cos u multiplied by the derivative of u with respect to x:
d/dx(sin u) = cos u * du/dx
In this case, u = 6x, so the derivative of sin 6x is cos 6x * d/dx(6x).
Using the power rule for the second term, we have:
d/dx(sin^2 6x) = 2 * sin 6x * cos 6x * d/dx(6x)
Combining these results, the derivative of y with respect to x is:
dy/dx = cos 6x * 6 + 2 * sin 6x * cos 6x * 6
Simplifying this expression, we have:
dy/dx = 6(cos 6x + 2sin 6x * cos 6x)
Step 2: Evaluate the derivative at the given point (0, 0).
Substitute x = 0 into the derivative:
dy/dx = 6(cos 0 + 2sin 0 * cos 0)
dy/dx = 6(1 + 0)
dy/dx = 6
The derivative of y with respect to x at x = 0 is 6.
Step 3: Use the slope-intercept form to find the equation of the tangent line.
The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept.
We already know the slope (m) of the tangent line, which is 6.
To find the y-intercept (b), substitute the coordinates of the given point (0, 0) into the equation:
0 = 6(0) + b
0 = b
Hence, the y-intercept is 0.
Therefore, the equation of the tangent line to the curve y = sin 6x + sin^2 6x at the point (0, 0) is:
y = 6x