now, make x=PI/6, and calculate f'
Then, y=f'(PI/6)*x + b
let f(x)= 6sin(x)/(2sin(x)+6cos(x))
The equation of the tangent line to y= f(x) at a= pi/6 can be written in tthe form y=mx+b where m=? b=?
I found the f':
(9*(cos^2(x)+sin^2(x)))/(3cos(x)+sin(x))^2
but i don't know how to fill it into the form. Can anyone help me?
Thanks you
2 answers
your f ' (x) is correct but not complete
f ' (x) = (9*(cos^2(x)+sin^2(x)))/(3cos(x)+sin(x))^2
remember that sin^2 x + cos^2 x = 1
f '(x) = 9/(3cosx + sinx)^2
so when x = π/6 ,
f'(π/6) = 9/(3√3/2 + 1/2)^2
= appr .93769
this is the slope of the tangent when x = π/6
when x = π/6
f(π/6) = 3/(1+3√3) = appr .4842
equation of tangent:
y - .4842 = .93769(x - .5236)
check my arithmetic
f ' (x) = (9*(cos^2(x)+sin^2(x)))/(3cos(x)+sin(x))^2
remember that sin^2 x + cos^2 x = 1
f '(x) = 9/(3cosx + sinx)^2
so when x = π/6 ,
f'(π/6) = 9/(3√3/2 + 1/2)^2
= appr .93769
this is the slope of the tangent when x = π/6
when x = π/6
f(π/6) = 3/(1+3√3) = appr .4842
equation of tangent:
y - .4842 = .93769(x - .5236)
check my arithmetic
2 answers