logarithmic differentiation

[MATH]y = x^{1/2} \cdot e^{x^2-x} \cdot (x+1)^{3/2}[/MATH]
[MATH]\log{y} = \dfrac{1}{2}\log{x} + (x^2-x) + \dfrac{3}{2}\log(x+1)[/MATH]
continue ...
 
Actually, in calculus it's far better to use \(\ln\), or \(\log_e\) -- whatever you choose to call it! And that is what was intended.

I apologize for using log as base e without stating such ... force of habit.
 
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