Asked by Anonymous
Evaluate the limit as h -> 0 of:
[tan (pi/6 + h) - tan(pi/6)]/h
I thought the answer was √3/3, or tan(pi/6, but apparently that is wrong, any tips here?
[tan (pi/6 + h) - tan(pi/6)]/h
I thought the answer was √3/3, or tan(pi/6, but apparently that is wrong, any tips here?
Answers
Answered by
Reiny
the basic definition of the derivative of any function f(x) is
Lim [f(x+h) - f(x)]/h as h --> 0
this is exactly the pattern we are looking at.
So what they are asking for is the derivative of
tan x when x = pi/6
let y = tan x
then dy/dx = sec^2 x
= sec^2 (pi/6) or sec^2 30º
= 4/3
( cos 30º = √3/2
sec 30º = 2/√3
sec ^2 30º = 4/3 )
Lim [f(x+h) - f(x)]/h as h --> 0
this is exactly the pattern we are looking at.
So what they are asking for is the derivative of
tan x when x = pi/6
let y = tan x
then dy/dx = sec^2 x
= sec^2 (pi/6) or sec^2 30º
= 4/3
( cos 30º = √3/2
sec 30º = 2/√3
sec ^2 30º = 4/3 )
Answered by
Anonymous
Thanks, I just forgot to do the dy/dx of tan x.
Answered by
Anonymous
12 year old thread but still saved my grade Thanks
Answered by
charlie
did not save my grade this is unreadable
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