C4 DifferentationWatch

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#1
I have been doing some C4 practice papers and there is the one question that I cannot get the right answer for.

Find the value of f'(x) at the value x=Pi/3 for f(x)=tan2xsinx

Now, I can get so far as getting f'(x) and according to the answer booklet it is correct at f'(x)=2.sec^2.2x.sinx + tan2x.cosx but when I substitute in Pi/3 I keep coming up with a wrong answer.

Please could someone help. Thank you
0
13 years ago
#2
f'(x)=2.sec^2.2x.sinx + tan2x.cosx when x=pi/3
= 2 x 4 x 0.866 + -1.732 x 0.5
=6.928 - 0.866
=6.06 (3sf)

Is this either of the answers you have?
0
13 years ago
#3
r u sure u set ur calculator on "radians"?
0
13 years ago
#4
(Original post by franks)
f'(x)=2.sec^2.2x.sinx + tan2x.cosx when x=pi/3
= 2 x 4 x 0.866 + -1.732 x 0.5
=6.928 - 0.866
=6.06 (3sf)

Is this either of the answers you have?
Seems to be correct
0
#5
Yes that is what I have been getting but the answers at the back say that it is wrong. Nevermind, I will put it as that. Thank you.
0
13 years ago
#6
(Original post by BS-E-M-Student)
Yes that is what I have been getting but the answers at the back say that it is wrong. Nevermind, I will put it as that. Thank you.
0
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