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farukk

Find dy\dx if y=(x+1)^x

  • one year ago
  • one year ago

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  1. RolyPoly
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    \[y=(x+1)^x\]\[lny = xln(x+1)\]Diff. both sides w.r.t. x

    • one year ago
  2. farukk
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    plz do it for me

    • one year ago
  3. farukk
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    i have tried but failed in getting answer

    • one year ago
  4. RolyPoly
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    Can you show your work?

    • one year ago
  5. farukk
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    i have a few questions similar to it so explain also

    • one year ago
  6. farukk
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    yeah

    • one year ago
  7. farukk
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    |dw:1355104426640:dw|

    • one year ago
  8. farukk
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    |dw:1355104863027:dw|

    • one year ago
  9. farukk
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    did u use this?

    • one year ago
  10. RolyPoly
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    \[y=(x+1)^x\]\[\ln y=x\ln(x+1)\]Diff. both sides w.r.t. x \[\frac{1}{y} \frac{dy}{dx} = x\frac{d}{dx}(\ln (x+1))+ ln(x+1)\frac{d}{dx}x\] Yes, product rule.

    • one year ago
  11. farukk
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    okay

    • one year ago
  12. robtobey
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    Refer to the attachment.

    • one year ago
    1 Attachment
  13. farukk
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    thanks all

    • one year ago
  14. farukk
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    @RolyPoly

    • one year ago
  15. RolyPoly
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    \[\frac{1}{y} \frac{dy}{dx} = x\frac{d}{dx}(\ln (x+1))+ ln(x+1)\frac{d}{dx}x\]\[\frac{1}{y} \frac{dy}{dx} = x(\frac{1}{x+1})+ ln(x+1)(1)\]\[\frac{1}{y} \frac{dy}{dx} = (\frac{x}{x+1})+ ln(x+1)\]\[\frac{dy}{dx} = y[(\frac{x}{x+1})+ ln(x+1)]\]Then sub y= (x+1)^x into dy/dx

    • one year ago
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