Find an equation of the tangent line to the curve y=sin(3x)+cos(4x) at the point (π/6,y(π/6))

Answer 1

Step 1: Take the derivative - that will give you a slope function that will allow you to find the slope at any value of x.

y' = 3cos(3x) - 4sin(4x)

Step 2: Plug in the value of x you want to find the slope at (in this case x = π/6):

y'(π/6) = 3cos(3•π/6) - 4sin(4•π/6) = 3cos(π/2) - 4sin(2π/3) = 3(0) - 4(√3/2) = -2√3

Step 3: Using the point-slope form of a line, plug in "m" and the point (x₁, y₁) to get the equation of the line.

In this case we need to calculate y(π/6) to get the point (x₁, y₁):

y = sin(3x) + cos(4x)

y = sin(3•π/6) + cos(4•π/6) = sin(π/2) + cos(2π/3) = 1 + -1/2 = 1/2

So the point (x₁, y₁) is (π/6, 1/2)

Since the point-slope form of a line is y - y₁ = m(x - x₁) the equation of the line is:

y - 1/2 = -2√3(x - π/6)