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anonymous

  • one year ago

Guys I give medals help me find the y' of y=cos(x+y)

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  1. anonymous
    • one year ago
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    chain rule again for this one, in the guess of "implicit diff"

  2. anonymous
    • one year ago
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    think of it as \[f(x)=\cos(x+f(x))\]

  3. anonymous
    • one year ago
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    then via the chain rule you get \[f'(x)=-\sin(x+f(x))\times \left(1+f'(x)\right)\] solve this equation for \(f'(x)\)

  4. anonymous
    • one year ago
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    of course it is easier to write \[y'=-\sin(x+y)(1+y')\] and solve for \(y'\)

  5. anonymous
    • one year ago
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    yeah, can I divide the y'+1 from one side to the other?

  6. anonymous
    • one year ago
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    ok I got it

  7. anonymous
    • one year ago
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    i will leave the algebra to you , but no, that is not how you solve you need to distribute first

  8. anonymous
    • one year ago
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    I got |dw:1439515086039:dw|

  9. anonymous
    • one year ago
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    i just remembered my algebra stuff.. thank you again

  10. anonymous
    • one year ago
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    looks good to me

  11. anonymous
    • one year ago
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    yeah just because you are taking calc doesn't mean the algebra has changed any if my experience algebra is the biggest pitfall, not the ideas of calculus

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spraguer (Moderator)
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