Find an equation of the tangent line to the given curve at the given point
y=(1/cosx)-2cosx at((pi/3),1)
y'=0+2sinx
slope = √3
y=mx+b
1=(√3)(pi/3)+b
b=(3-√3pi)/3
3√3x-3y+3-√3pi=0
What did I do wrong? Thanks in advance.
Got it.
To find the equation of the tangent line to the curve at the given point, you first need to find the derivative of the function. Let's go through the steps again:
Given function: y = (1/cosx) - 2cosx
1. Find the derivative of y with respect to x (y') by using the quotient rule and chain rule:
y' = [(-sinx)(cosx) - (1/cosx)(-sinx)]/ cos^2x
= [-sin^2x + sinx/cosx]/ cos^2x
= [sinx - sin^2x/cosx]/ cos^2x
= (sinx (1 - sinx))/cos^3x
2. Now, we need to find the value of y' at the given point (π/3, 1):
Plug in x = π/3 into the derived equation to find y':
y'(π/3) = (sin(π/3)(1 - sin(π/3)))/cos^3(π/3)
= [(√3/2)(1 - √3/2)]/(1/2)^3
= [(√3 - 3/2)/(1/8)]
= [8(√3 - 3)/2]
= 4(√3 - 3)
= 4√3 - 12
3. Now that we have the slope of the tangent line, we can use the point-slope form of the equation of a line:
y - y1 = m(x - x1)
y - 1 = (4√3 - 12)(x - π/3)
Distribute:
y - 1 = 4√3x - 4√3(π/3) - 12x + 4(π/3)
y = (4√3 - 12)x - 4√3(π/3) + 4(π/3) + 1
Simplify the constants:
y = (4√3 - 12)x - 4√3π/3 + 4π/3 + 1
y = (4√3 - 12)x + (-4√3π + 4π + 3)/3
So, the equation of the tangent line to the given curve at the point (π/3, 1) is y = (4√3 - 12)x + (-4√3π + 4π + 3)/3.
I hope this clears up the process for finding the equation of the tangent line. Let me know if you have any further questions!