If f(x)=tan(2x), then f'(pi/6)=?

Question

(a) Square root of 3
(b) 2 * square root of 3
(c) 4
(d) 4 * square root of 3
(e) 8

Answer 1

f(x) = tan(2x)

We differentiate this, keeping the chain rule in mind.

f'(x) = [sec^2(2x)] (2), which can be simplified as
f'(x) = 2sec^2(2x)

Now, we plug in x = pi/6 for f'(x).

f'(pi/6) = 2sec^2(2 * pi/6)

Simplifying the terms inside the brackets give us (pi/3).

f'(pi/6) = 2sec^2(pi/3)

Remember that sec = 1/cos; therefore,

f'(pi/6) = 2/cos^2(pi/3)

Remember that cos^2(x) is defined to be [cos(x)]^2. Therefore,

f'(pi/6) = 2/[cos(pi/3)]^2

We know what cos(pi/3) is; it's one of our known values on the unit circle. It's equal to 1/2. Therefore,

f'(pi/6) = 2/[1/2]^2, which is the same as
f'(pi/6) = 2/[1/4], which is a complex fraction. Multiplying numerator and denminator by 4 gives us

f'(pi/6) = 2(4)/1 = 8

Answer 2

If f(x)=tan(2x), then f'(pi/6)=?
f(pi/6) = tan(2*pi/6) = tan pi/3 = square root of 3
pi/3 is 60 degrees. In a 30-60-90 degree triangle the hyptenuse is 2, the short leg is 1 and the long leg is the sqrt(3)
So tan 60 degrees = lonb leg /short leg = 3qrt(3)/1 = sqrt(3)

Answer 3

I don't know where people are getting 8 from, but if you plug x=PI/6 into tan(2PI/6)=tan(PI/3)=sin(PI/3)/cos(PI/3)=((root3/2)/(1/2))= root 3
So the answer is (a).

You can also use the double angle formula to solve this. (2 tan x)/(1-tan^2x)

oops, is that suppose to be f prime? didn't catch that.

2sec^2(PI/3)=2/cos^2(PI/3)=8

Hehe sorry about that.

Answer 4

f(x)=tan(2x)

f'(x) = 2 sec² (2x)

So f'(π/6) = 2 sec² (π/3)

= 2/cos²(π/3)

= 2/(½)²

= 2/¼

= 8 ← option (e)

Answer 5

f(x)=tan2x
f'(x)=2sec^2(2x)
f'(pi/6)=2*sec^2(pi/3)
=2[(sec pi/3)^2]
=2*2^2
=8

Answer 6

e ans is 8