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farukk

  • 2 years ago

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

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

  2. farukk
    • 2 years ago
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    plz do it for me

  3. farukk
    • 2 years ago
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    i have tried but failed in getting answer

  4. RolyPoly
    • 2 years ago
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    Can you show your work?

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

  6. farukk
    • 2 years ago
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    yeah

  7. farukk
    • 2 years ago
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    |dw:1355104426640:dw|

  8. farukk
    • 2 years ago
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    |dw:1355104863027:dw|

  9. farukk
    • 2 years ago
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    did u use this?

  10. RolyPoly
    • 2 years ago
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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.

  11. farukk
    • 2 years ago
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    okay

  12. robtobey
    • 2 years ago
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    Refer to the attachment.

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  13. farukk
    • 2 years ago
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    thanks all

  14. farukk
    • 2 years ago
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    @RolyPoly

  15. RolyPoly
    • 2 years ago
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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

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