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C3 Differentiation

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! :smile:
To get it in terms of x you mean? Use trig identities or compare the sides of a triangle.
(edited 7 years ago)
Reply 2
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solC
14
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
Reply 3
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solC
14
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 acosec(3y)acosec(3y).


I think the answer given is correct, since dx/dy should simplify to 3sin6y.
Original post by solC
I think the answer given is correct, since dx/dy should simplify to 3sin6y.



6sin3y, because of the chain rule.
Reply 5
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solC
14
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:s-smilie:
Original post by solC
I got 3(2)sin(3y)cos(3y) which is the same as 3sin6y?
Maybe i'm missing something:s-smilie:


Oh yeah, you're right. Enough maths for the day for me.
Reply 7
Avatar for Zacken
Zacken
22
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! :smile:


Differentiate, like you said, you get:

dxdy=3sin⁡6y\frac{dx}{dy} = 3 \sin 6y, now remember that you want dydx\frac{dy}{dx}, so flip both sides of your equation over to get dydx=13sin⁡6y=13×1sin⁡6y\frac{dy}{dx} = \frac{1}{3\sin 6y} = \frac{1}{3} \times \frac{1}{\sin 6y}

But you know that cosec a=1sin⁡a\mathrm{cosec} \, a = \frac{1}{\sin a} so here cosec 6y=1sin⁡6y\mathrm{cosec} \, 6y = \frac{1}{\sin 6y}, so your derivative is 13 cosec 6y\frac{1}{3} \, \mathrm{cosec} \, 6y
Original post by Zacken
Differentiate, like you said, you get:

dxdy=3sin⁡6y\frac{dx}{dy} = 3 \sin 6y, now remember that you want dydx\frac{dy}{dx}, so flip both sides of your equation over to get dydx=13sin⁡6y=13×1sin⁡6y\frac{dy}{dx} = \frac{1}{3\sin 6y} = \frac{1}{3} \times \frac{1}{\sin 6y}

But you know that cosec a=1sin⁡a\mathrm{cosec} \, a = \frac{1}{\sin a} so here cosec 6y=1sin⁡6y\mathrm{cosec} \, 6y = \frac{1}{\sin 6y}, so your derivative is 13 cosec 6y\frac{1}{3} \, \mathrm{cosec} \, 6y


Alright. :smile:
Thank you for your time :smile:
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.
(edited 7 years ago)
Reply 9
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Zacken
22
Original post by sabahshahed294
Alright. :smile:
Thank you for your time :smile:
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 x=sin⁡23yx = \sin^2 3y, now you're going to differentiate with respect to y. (which is why it's dy at the bottom)

That gets you
Unparseable latex formula:

\frac{dx}{dy} = 2 \times \sin^{2-1} 3y \times 3\cos 3y = 6 \times \sin 3y \ties \cos 3y

by the chain rule. Then trig identity.

Remember if you have to differentiate [f(y)]n[f(y)]^n with respect to yy, that gets you n[f(y)]n−1×f′(y)n[f(y)]^{n-1} \times f'(y) which is exactly what you have here with n=2n=2 and f(y)=sin⁡3yf(y) = \sin 3y.
(edited 7 years ago)
Original post by Zacken
Nope, you've started out with x=sin⁡23yx = \sin^2 3y, now you're going to differentiate with respect to y. (which is why it's dy at the bottom)

That gets you
Unparseable latex formula:

\frac{dx}{dy} = 2 \times \sin^{2-1} 3y \times 3\cos 3y = 6 \times \sin 3y \ties \cos 3y

by the chain rule. Then trig identity.

Remember if you have to differentiate [f(y)]n[f(y)]^n with respect to yy, that gets you n[f(y)]n−1×f′(y)n[f(y)]^{n-1} \times f'(y) which is exactly what you have here with n=2n=2 and f(y)=sin⁡3yf(y) = \sin 3y.


Oh. Right. My bad! Thanks :smile:
Reply 11
Avatar for Zacken
Zacken
22
Original post by sabahshahed294
Oh. Right. My bad! Thanks :smile:


You're welcome.
Original post by Zacken
Nope, you've started out with x=sin⁡23yx = \sin^2 3y, now you're going to differentiate with respect to y. (which is why it's dy at the bottom)

That gets you
Unparseable latex formula:

\frac{dx}{dy} = 2 \times \sin^{2-1} 3y \times 3\cos 3y = 6 \times \sin 3y \ties \cos 3y

by the chain rule. Then trig identity.

Remember if you have to differentiate [f(y)]n[f(y)]^n with respect to yy, that gets you n[f(y)]n−1×f′(y)n[f(y)]^{n-1} \times f'(y) which is exactly what you have here with n=2n=2 and f(y)=sin⁡3yf(y) = \sin 3y.


it is much easier to rewrite **x=sin⁡23yx = \sin^2 3y, as { 1 - Cos6y }/2 then differentiate wrt y

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