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# Can you Differentiate tan2 theta !? watch

1. Parametric Equation

x= tan theta, y= tan2theta

Find the value if dy/dx when theta= pi/6

I differentiated tan theta got 1/cos theta which is sec? but stuck on the next part

Thx ..
2. Try the chain rule
3. let 2theta = x
then differentiate tan x and then sub in the thetas in at the end.
4. (Original post by mufas)
Parametric Equation

x= tan theta, y= tan2theta

Find the value if dy/dx when theta= pi/6

I differentiated tan theta got 1/cos theta which is sec? but stuck on the next part

Thx ..
5. (Original post by Ben121)
Try the chain rule
got 2sec^2x but still dont work
6. (Original post by EEngWillow)
this one
7. (Original post by mufas)
this one
In that case use the chain rule with u = 2theta as suggested. Post your working if you got the wrong answer and we can help you further.

EDIT: Just saw your post above. The argument inside your trig function should still be 2theta. Otherwise you've done it correctly.
8. (Original post by EEngWillow)
In that case use the chain rule with u = 2theta as suggested. Post your working if you got the wrong answer and we can help you further.

EDIT: Just saw your post above. The argument inside your trig function should still be 2theta. Otherwise you've done it correctly.
Again I got 2/cos2theta

so when (2/cos2theta) / (1/cos theta )

= 2cos theta / cos 2 theta

when i sub pi/6 into the top one i get decimals , but the answer is 6!
9. (Original post by mufas)
Again I got 2/cos2theta

so when (2/cos2theta) / (1/cos theta )

= 2cos theta / cos 2 theta

when i sub pi/6 into the top one i get decimals , but the answer is 6!
Ah, I've seen your problem. Silly me. The derivitive of tan(theta) isn't sec(theta). Try doing that one again.
10. (Original post by EEngWillow)
Ah, I've seen your problem. Silly me. The derivitive of tan(theta) isn't sec(theta). Try doing that one again.
checked again same ans whats wrong?

its 2/cos2theta ?
11. try differentiating tan with the quotient rule, remembering tanx = (sinx)/(cosx) and see what the standard result for differentiating tan is
12. (Original post by RK92)
try differentiating tan with the quotient rule, remembering tanx = (sinx)/(cosx) and see what the standard result for differentiating tan is
its tan 2 theta not just tan theta
13. Its been 40 min still no solution !! I'm still trying!
14. (Original post by mufas)
checked again same ans whats wrong?

its 2/cos2theta ?
Basically,

which is what you put in the first post. It's what you've calculated for the very first part. I suspect this mistake was carried through into your calculation of which is why you're getting the wrong answer.
15. (Original post by EEngWillow)
Basically,

which is what you put in the first post. It's what you've calculated for the very first part. I suspect this mistake was carried through into your calculation of which is why you're getting the wrong answer.
I'm sorry I'm completely LOST can you please post The working for some bits at least, since i dont want to take any more of your time
16. u can....
17. Can somebody atleast give me the solution EVEN without no working, so that i could work towards it
18. (Original post by mufas)
its tan 2 theta not just tan theta
yeah, you can use the chain rule afterwards.

Spoiler:
Show

il walk you through it.
you want to differentiate u/v

you have u=sin(2x) and v=cos(2x)

quotient rule says the derivative is: (u'.v - u.v')/v²

which is: [ 2cos(2x).cos(2x) - -2sin(2x).sin(2x) ] / cos²(2x) = [ 2cos²(2x) + 2sin²(2x) ]/cos²(2x) = 2[ cos²(2x) + sin²(2x) ]/cos²(2x) = 2/cos²(2x) = 2sec²(2x)
19. (Original post by RK92)
yeah, you can use the chain rule afterwards.

Spoiler:
Show

il walk you through it.
you want to differentiate u/v

you have u=sin(2x) and v=cos(2x)

quotient rule says the derivative is: (u'.v - u.v')/v²

which is: [ 2cos(2x).cos(2x) - -2sin(2x).sin(2x) ] / cos²(2x) = [ 2cos²(2x) + 2sin²(2x) ]/cos²(2x) = 2[ cos²(2x) + sin²(2x) ]/cos²(2x) = 2/cos²(2x) = 2sec²(2x)
I have got this same answer earlier on but it doesn't work for some reason

(1/cosx)/(2sec²(2x) ) = 2cosx/ cos2x so when you sub the pi/6 you dont get 6
20. (Original post by mufas)
I have got this same answer earlier on but it doesn't work for some reason

(1/cosx)/(2sec²(2x) ) = 2cosx/ cos2x so when you sub the pi/6 you dont get 6
I'll repeat what I said then. In your first post, you've said this:

I differentiated tan theta got 1/cos theta
This is not correct. You need to fix this before you can get the right answer.

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