Logarithmic Differentiation — Question 9

Question 9 - Solution

For $x\ne\pm1$, the square root simplifies to

$$y=\frac{(x^2+4)^{3/2}|x-1|}{(x^2-1)^4}>0.$$

Hence

$$\ln y=\tfrac32\ln(x^2+4)+\ln|x-1|-4\ln|x^2-1|.$$

Differentiating and multiplying by $y$ gives

$$\boxed{y'=\frac{(x^2+4)^{3/2}|x-1|}{(x^2-1)^4} \left(\frac{3x}{x^2+4}+\frac1{x-1}-\frac{8x}{x^2-1}\right),\quad x\ne\pm1.}$$