Given that y = (1 +2x) ^ x , prove that dy/dx = y (ln (2x +1) +1 - (1/ (1+2x)) )

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- Aug 5th 2008, 03:51 AMTangeraDifferentiation
Given that y = (1 +2x) ^ x , prove that dy/dx = y (ln (2x +1) +1 - (1/ (1+2x)) )

Thank you for helping! - Aug 5th 2008, 04:00 AMflyingsquirrel
Hello !$\displaystyle y=(1+2x)^x=\exp\left[ x\ln(1+2x) \right]$. Let $\displaystyle u(x)=x\ln(1+2x)$. $\displaystyle y$ can now be written $\displaystyle y=\exp(u(x))$ and the chain rule states that $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=u'(x)\exp(u(x))$ : once you have computed $\displaystyle u'$, you are done. Does it help ?

- Aug 5th 2008, 06:09 AMTangera
Hello!

Yes, it helps a lot! Thank you for guiding me!