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1018

  • one year ago

find y' : y = sin (x+y) ( i have the answer but i need the solution thanks)

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  1. anonymous
    • one year ago
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    Implicit differentiation\[y = \sin \left( x + y \right)\]Differentiate both side wrt \(x\)\[\frac{ dy }{ dx } = \frac{ d }{ dx } \sin \left( x+y \right)\]Chain rule\[\frac{ dy }{ dx } = \cos \left( x + y \right) \left( 1 + \frac{ dy }{ dx } \right)\]Expand right hand side and solve for \(\frac{dy}{dx}\)

  2. 1018
    • one year ago
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    damn chain rule! thanks. hey can you show me how you would expand the right hand side? i was told i need to put all the dy/dx on one side etc etc. may you show me how would you do it thanks

  3. 1018
    • one year ago
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    and i have another question if you dont mind. haha. but it can be for later. haha

  4. anonymous
    • one year ago
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    Sure. \[\frac{ dy }{ dx } = \cos(x + y)\left( 1 + \frac{ dy }{ dx } \right)\]\[\frac{ dy }{ dx } = \cos \left( x + y \right) + \frac{ dy }{ dx }\cos \left( x + y \right)\]\[\frac{ dy }{ dx } - \frac{ dy }{ dx }\cos \left( x + y \right) = \cos \left( x + y \right)\]\[\frac{ dy }{ dx }\left( 1-\cos \left( x+y \right) \right) = \cos \left( x + y \right)\]Can you take it from here?

  5. 1018
    • one year ago
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    ooh! yes yes thanks i see it now. wait i have to ask, can i, in a way, 'adjust' the first equation given? or is it to be differentiated as it was?

  6. anonymous
    • one year ago
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    Not that I'm aware of. I think the problem is for you to practice implicit differentiation just as it is given.

  7. 1018
    • one year ago
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    ok. thanks again!

  8. anonymous
    • one year ago
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    You're welcome

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