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Math Help - Derivatives of inverse trig functions.

  1. #16
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    After you reach the point where you have \ln{y} = \:3x\cdot\ln(\sin 5x). Differentiate both sides. To do so, use
    the chain rule on the the left hand side: [\ln{y}]' = \frac{y'}{y}. That done, use the chain and product rules on the
    right hand side; I think you know how (it's the product of two functions one of which is a composite, just
    like the problem from the other thread). Having done that, multiply both sides by y, then you are done!
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  2. #17
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    I'm sorry I keep posting, I have trouble understanding unless I do it myself. Here is up to where I understood.



    Why does he multiply it again by the original function in the very end though (in his post)? And is there a more simplified way to write it down and how would I be able to reduce (because this specific problem is from the multiple choice section and the choices do not correspond with the answer).
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  3. #18
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    Quote Originally Posted by FullBox View Post
    Why does he multiply it again by the original function in the very end though (in his post)? And is there a more simplified way to write it down and how would I be able to reduce (because this specific problem is from the multiple choice section and the choices do not correspond with the answer).
    Because you took the logarithm of both sides, you no longer have y on the other side. You have ln(y), right?
    So usually while differentiating you get y on one side, but now you have [log(y)]', which is y'/y by the chain
    rule. So now, because you want to get your derivative in terms of y' = ... you multiply y on both sides. Thus:

    \frac{y'}{y} \;=\;15x\cot5x + 3\ln(\sin5x) \Rightarrow y'  \;=\;[15x\cot5x + 3\ln(\sin5x)]y

    You have few typos at the end of your post, but your work looks right.
    Last edited by TheCoffeeMachine; November 22nd 2010 at 10:16 PM.
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  4. #19
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    Oh okay, I think I get it now. And y = (sin5x)^3x in our case.

    But what do you mean when you say "you multiply y on both sides"? Do you mean on the y' side and the [15xcot5x + 3ln(sin5x)] side?

    And is there a way for us to simplify the answer further, or is that as far as we can get? The choices that the question provides, while they do resemble, do not correspond with the answer we have reached.

    Sorry I am very tired, I'm not sure if I'm even making sense anymore. Its 2:12 A.M. here and I need to go to sleep now. But I will be back tomorrow to hopefully continue on the parts I am not understanding.
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  5. #20
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    Quote Originally Posted by FullBox View Post
    Oh okay, I think I get it now. And y = (sin5x)^3x in our case.
    Yes!

    But what do you mean when you say "you multiply y on both sides"? Do you mean on the y' side and the [15xcot5x + 3ln(sin5x)] side?
    Well, of course. You have y'/y on one side, and you want to express your derivative explicitly -- i.e. y' = [something], instead of y'/y = [something].

    And is there a way for us to simplify the answer further, or is that as far as we can get? The choices that the question provides, while they do resemble, do not correspond with the answer we have reached.
    Would you be able to post the given choice? I can't see a more reasonable form.
    Sorry I am very tired, I'm not sure if I'm even making sense anymore. Its 2:12 A.M. here and I need to go to sleep now. But I will be back tomorrow to hopefully continue on the parts I am not understanding.
    It's late over here. Very, very late that the sun has already risen!
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  6. #21
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    By the way, you are consistently writing " \frac{dy}{dx} f(x)" where you mean \frac{d f(x)}{dx}. The "y" represents the function f(x) and so is redundant. If you cannot fit the entire function into the derivative symbol, use \frac{d}{dx}f(x).
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  7. #22
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    Quote Originally Posted by TheCoffeeMachine View Post
    Well, of course. You have y'/y on one side, and you want to express your derivative explicitly -- i.e. y' = [something], instead of y'/y = [something].
    Sorry, it confused me at first because, as HallsOfIvy pointed out, I was not writing it correctly, so I wasn't sure what that was at first. However, when I realized its simply dy/dx with y being f(x) it made a lot more sense.

    Would you be able to post the given choice? I can't see a more reasonable form.
    These were the choices given to me that I can pick from. Only one is right, but I cannot see how I would further simplify the answer we reached to achieve one of them.



    Quote Originally Posted by HallsofIvy View Post
    By the way, you are consistently writing " \frac{dy}{dx} f(x)" where you mean \frac{d f(x)}{dx}. The "y" represents the function f(x) and so is redundant. If you cannot fit the entire function into the derivative symbol, use \frac{d}{dx}f(x).
    Yeah, we are finding the derivative of the function f(x) with respect to x. I'm sorry I keep writing it incorrectly, I just forget sometimes. Like for example, I have made mistakes when we are trying to find the derivative of theta with respect to x, and instead of writing theta I accidentally write y. I tend to forget since the theorems are all written in dy/dx, so I sometimes mess up and forget to replace y with what its supposed to be.
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  8. #23
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    Quote Originally Posted by FullBox View Post
    \displaystyle 3\left(\sin{5x}\right)^{3x}\left(5x\cot{5x}+\ln{\s  in{5x}}\right) = \left(\sin{5x}\right)^{3x}\left[(3)(5x)\cot{5x}+(3)\ln{\sin{5x}}\right] \displaystyle = \left(\sin{5x}\right)^{3x}\left(15x\cot{5x}+3\ln{\  sin{5x}}\right).
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