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DLS

  • 3 years ago

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  1. DLS
    • 3 years ago
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    \[xcosy+ycosx=\tan^{-1}x^{2}\]

  2. DLS
    • 3 years ago
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    this is how i did it

  3. DLS
    • 3 years ago
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    \[\frac{d(xcosy)}{dx} + \frac{d(ycosx)}{dx} = d(\tan^{-1} x^{2})\]

  4. DLS
    • 3 years ago
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    \[cosy-xsin \frac{dy}{dx} + \dfrac{(ycosx)}{dx} - ysinx=\frac{1}{1+x^{4}}\]

  5. DLS
    • 3 years ago
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    \[\frac{dy}{dx}(cosx-xsiny)=\frac{1}{1+x^{4}}-cosy+ysinx\]

  6. DLS
    • 3 years ago
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    \[\frac{dy}{dx}=\frac{1}{1+x^{4}(cosy-xsiny}) - \frac{cosy+ysinx}{cosx-xsiny}\]

  7. DLS
    • 3 years ago
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    2nd equation is wrong sorry

  8. hartnn
    • 3 years ago
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    \(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosy-xsiny)}- \frac{cosy-ysinx}{cosx-xsiny}\)

  9. DLS
    • 3 years ago
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    \[cosy-xsiny \frac{dy}{dx} +cosx \frac{dy}{dx}-ysinx=\frac{1}{1+x^{4}}\]

  10. DLS
    • 3 years ago
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    oh so mine is correct? :O

  11. hartnn
    • 3 years ago
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    what u wrote now is correct.

  12. DLS
    • 3 years ago
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    finalans is correct right

  13. hartnn
    • 3 years ago
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    u get this \(\large \frac{dy}{dx}=\frac{1}{(1+x^{4})(cosy-xsiny)}- \frac{cosy-ysinx}{cosx-xsiny}\)

  14. hartnn
    • 3 years ago
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    cos y - y sin x

  15. DLS
    • 3 years ago
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    i got that only

  16. hartnn
    • 3 years ago
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    u got cos y+ y sin x

  17. DLS
    • 3 years ago
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    ah minor mistakes :p

  18. DLS
    • 3 years ago
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    if u open ur bracket u wud still get -cosy+ysinx i guess

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