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anonymous

  • one year ago

Find dy/dx of x cos(y) = 1 at the point (-1,pi) using implicit differentiation

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  1. anonymous
    • one year ago
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    I solved this problem on a test and got it right, but now I don't remember how to do it. Find \[\frac{ dy }{ dx } x \cos(y)\] at the point (−1,π)

  2. dessyj1
    • one year ago
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    It seems like we will have to use the product rule. Are you familiar with the product rule?

  3. nincompoop
    • one year ago
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    do you want to just get the general derivative implicitly first then add the values later

  4. anonymous
    • one year ago
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    Yes. But do I keep the x as it is or do I rplacee it with -1?

  5. nincompoop
    • one year ago
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    product rule, kid

  6. anonymous
    • one year ago
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    @nincompoop I know the product rule

  7. dessyj1
    • one year ago
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    Do you need help with implicitly deriving the left side of your equation?

  8. anonymous
    • one year ago
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    I know ow to do the implicit part. This is what I did on the test (and got it correct but now it doesn't make any sense) Step 1: (-1) (-sin(y))y' + cos(y) (-1) = 0 Step 2: sin(y) y' - cos(y) Step 3: sin(y) y' = cos(y) Step 4: \[y' = \frac{ \cos(y) }{ \sin(y) }\] In step 1 shouldn't it be + cos(y) (0) ? but I didn't write that and still got the answer correct

  9. dessyj1
    • one year ago
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    It would not be cos(y)(0) because the derivative of x is 1

  10. dessyj1
    • one year ago
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    Also, of you were following the f'g +fg' you will realize that you forgot to multiply y'sin(y) by an x.

  11. dessyj1
    • one year ago
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    Remember that the derivative of a constant is 0

  12. anonymous
    • one year ago
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    Is this correct? cos(y) (1) + x (-sin(y) y' = 0 cos(y) - x sin(y) y' = 0 cos(y) = x sin(y) y' \[y' = \frac{ \cos(y) }{ x \sin (y) }\] \[y' = \frac{ \cos(y) }{ x \sin (y) } = \frac{ \cos \pi }{ (-1) \sin \pi } = \frac{ -1 }{ (-1)(0) }\] Hence it is undefined

  13. dessyj1
    • one year ago
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    The math looks correct, but I do not think it would be undefined, rather it would not exist.

  14. anonymous
    • one year ago
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    @dessyj1 Thank you

  15. dessyj1
    • one year ago
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    np

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