(f'(π/3) = (-π/3 * √3/2 - 1/2)/(π/3)^2
= -(9π-3√3)/(2π^2)
If we call that slope m, then the tangent line is
y-1 = m(x-π/3)
Differentiate.
f(x)=(cos x)/x
I got to here:
(-x sin x - cos x)/x^2
Find the equation of the tangent line to
f(x)=sec x - 2 cos x at (pi/3, 1).
Is the slope 3 times sqrt of 3?
Equation is y=3 times sqrt of 3x - pi times sqrt of 3 +1??
1 answer