# Differentiation

• Aug 5th 2008, 03:51 AM
Tangera
Differentiation
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 AM
flyingsquirrel
Hello !
Quote:

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

$y=(1+2x)^x=\exp\left[ x\ln(1+2x) \right]$. Let $u(x)=x\ln(1+2x)$. $y$ can now be written $y=\exp(u(x))$ and the chain rule states that $\frac{\mathrm{d}y}{\mathrm{d}x}=u'(x)\exp(u(x))$ : once you have computed $u'$, you are done. Does it help ?
• Aug 5th 2008, 06:09 AM
Tangera
Hello!

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