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

Mathematics
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\[y=(x+1)^x\]\[lny = xln(x+1)\]Diff. both sides w.r.t. x
plz do it for me
i have tried but failed in getting answer

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Can you show your work?
i have a few questions similar to it so explain also
yeah
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did u use this?
\[y=(x+1)^x\]\[\ln y=x\ln(x+1)\]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.
okay
Refer to the attachment.
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thanks all
\[\frac{1}{y} \frac{dy}{dx} = x\frac{d}{dx}(\ln (x+1))+ ln(x+1)\frac{d}{dx}x\]\[\frac{1}{y} \frac{dy}{dx} = x(\frac{1}{x+1})+ ln(x+1)(1)\]\[\frac{1}{y} \frac{dy}{dx} = (\frac{x}{x+1})+ ln(x+1)\]\[\frac{dy}{dx} = y[(\frac{x}{x+1})+ ln(x+1)]\]Then sub y= (x+1)^x into dy/dx

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