Help with calculus?

Question

How would you find the equation of the tangent line to this equation at the point (pi/3, 1): 2cos^3(x). This is read as two times the cos of x cubed(I think...) The 3 is raised as an exponent between the cos and the x if that helps.

Answer 1

pi/3 is a #=(3.14.../3)
2cos^3(x)= cosx times cosx times cosx times 2
hope that helps

Answer 2

Take the derivate of F(x) = 2cos^3(x)

F'(x) = -2sin(x) cos^2(x)

To find the slope solve F'(pi/3)

F'(pi/3) = -2 (sqrt[3]/2) (1/2)^2 = - (sqrt 3)/4

Then solve using simple y=mx+b substitution

1 = [ - (sqrt 3)/4 ](pi/3) + b

b = 1 + (pi * sqrt 3)/12

So therefore, the equation of the tangent line is:

y = - (sqrt 3)/4 * x + 1 + (pi * sqrt 3)/12

Answer 3

if your equation is y = 2. cos^3(x)
then I doubt if the point (π/3, 1) lies on the curve. And if the point is not on the curve I don't think you can find the tangent to the curve at that point.
however to find the slope of the curve when x = π/3
differentiate the equation with respect to x

d(2.cos(x).cos(x).cos(x))
−
dx

And substitute value of x in the equation. this slope is also the slope of the tangent.

Answer 4

Take the deriv. of the equation:

f'(x) = 6cos^2(x)*-sinx = -6sin(x)[cos(x)]^2

Then replace x with pi/3 to get your slope:

m = -6(sqrt3/2)(1/2)^2 = -3sqrt3/4

Finally, go to the point-slope eqn of a line, plugging in (pi/3, 1) and your slope.

y-1 = -3sqrt3/4(x-pi/3)

Simplify if required.

Answer 5

take the derivative ... although maybe you know that