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# C3 Differentiation

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1. Hi, if we're given this function:-

x= sin2 3y and we are told to find dy/dx , the answer comes as 1/3 cosec 6y. My question is that how do we exactly produce this sort of an answer? I get that we differentiate firstly the function with respect to y and then take the reciprocal of dx/dy but I tend to get confused after this part actually i.e. after taking the reciprocal. Could somebody clear it out? TIA!
2. To get it in terms of x you mean? Use trig identities or compare the sides of a triangle.
3. If you first differentiate w.r.t. x you should be able to use a trig identity (double angle) to simplify and then take reciprocal. remember cosecx=1/sinx
4. (Original post by NotNotBatman)
To get it in terms of x you mean? Use trig identities or compare the sides of a triangle. dy/dx Is not 1/3 cosec 6y , it should be .
I think the answer given is correct, since dx/dy should simplify to 3sin6y.
5. (Original post by solC)
I think the answer given is correct, since dx/dy should simplify to 3sin6y.

6sin3y, because of the chain rule.
6. (Original post by NotNotBatman)
6sin3y, because of the chain rule.
I got 3(2)sin(3y)cos(3y) which is the same as 3sin6y?
Maybe i'm missing something
7. (Original post by solC)
I got 3(2)sin(3y)cos(3y) which is the same as 3sin6y?
Maybe i'm missing something
Oh yeah, you're right. Enough maths for the day for me.
8. (Original post by sabahshahed294)
Hi, if we're given this function:-

x= sin2 3y and we are told to find dy/dx , the answer comes as 1/3 cosec 6y. My question is that how do we exactly produce this sort of an answer? I get that we differentiate firstly the function with respect to y and then take the reciprocal of dx/dy but I tend to get confused after this part actually i.e. after taking the reciprocal. Could somebody clear it out? TIA!
Differentiate, like you said, you get:

, now remember that you want , so flip both sides of your equation over to get

But you know that so here , so your derivative is
9. (Original post by Zacken)
Differentiate, like you said, you get:

, now remember that you want , so flip both sides of your equation over to get

But you know that so here , so your derivative is
Alright.
Another thing, when differentiating this function with respect to x, it goes like dx/dy= 2 sin 3y multiplied by 3 cos 3y, right? As you take the derivative of the base and the angle. Sorry I'm kinda weak in this part of C3.
10. (Original post by sabahshahed294)
Alright.
Another thing, when differentiating this function with respect to x, it goes like dx/dy= 3 sin 6y multiplied by cos 6y, right? Sorry I'm kinda weak in this part of C3.
Nope, you've started out with , now you're going to differentiate with respect to y. (which is why it's dy at the bottom)

That gets you by the chain rule. Then trig identity.

Remember if you have to differentiate with respect to , that gets you which is exactly what you have here with and .
11. (Original post by Zacken)
Nope, you've started out with , now you're going to differentiate with respect to y. (which is why it's dy at the bottom)

That gets you by the chain rule. Then trig identity.

Remember if you have to differentiate with respect to , that gets you which is exactly what you have here with and .
12. (Original post by sabahshahed294)
You're welcome.
13. (Original post by Zacken)
Nope, you've started out with , now you're going to differentiate with respect to y. (which is why it's dy at the bottom)

That gets you by the chain rule. Then trig identity.

Remember if you have to differentiate with respect to , that gets you which is exactly what you have here with and .
it is much easier to rewrite **, as { 1 - Cos6y }/2 then differentiate wrt y

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