Asked by Anonymous
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
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
Answers
Answered by
bobpursley
now, make x=PI/6, and calculate f'
Then, y=f'(PI/6)*x + b
Then, y=f'(PI/6)*x + b
Answered by
Reiny
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.