ln(x(x^2-1)^1/2), I tried, now I have splinters Send help!
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Originally Posted by Neversh ln(x(x^2-1)^1/2), I tried, now I have splinters Send help! Maybe in this case it is easiest to first apply a little transformation, like this $\displaystyle \Big(\ln(x(x^2-1)^{1/2})\Big)'=\left(\ln x+\frac{1}{2}\ln(x^2-1)\right)'=\frac{1}{x}+\frac{1}{2}\frac{1}{x^2-1}\cdot 2x=\ldots$
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