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    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!
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    To get it in terms of x you mean? Use trig identities or compare the sides of a triangle.
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    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
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    (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).
    I think the answer given is correct, since dx/dy should simplify to 3sin6y.
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    (Original post by solC)
    I think the answer given is correct, since dx/dy should simplify to 3sin6y.

    6sin3y, because of the chain rule.
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    (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
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    (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.
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    (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:

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

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

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

    But you know that \mathrm{cosec} \, a = \frac{1}{\sin a} so here \mathrm{cosec} \, 6y = \frac{1}{\sin 6y}, so your derivative is \frac{1}{3} \, \mathrm{cosec} \, 6y
    Alright.
    Thank you for your time
    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.
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    (Original post by sabahshahed294)
    Alright.
    Thank you for your time
    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^2 3y, now you're going to differentiate with respect to y. (which is why it's dy at the bottom)

    That gets you \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 with respect to y, that gets you n[f(y)]^{n-1} \times f'(y) which is exactly what you have here with n=2 and f(y) = \sin 3y.
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    (Original post by Zacken)
    Nope, you've started out with x = \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 \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 with respect to y, that gets you n[f(y)]^{n-1} \times f'(y) which is exactly what you have here with n=2 and f(y) = \sin 3y.
    Oh. Right. My bad! Thanks
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    (Original post by sabahshahed294)
    Oh. Right. My bad! Thanks
    You're welcome.
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    (Original post by Zacken)
    Nope, you've started out with x = \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 \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 with respect to y, that gets you n[f(y)]^{n-1} \times f'(y) which is exactly what you have here with n=2 and f(y) = \sin 3y.
    it is much easier to rewrite **x = \sin^2 3y, as { 1 - Cos6y }/2 then differentiate wrt y
 
 
 
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