Logarithmic Differentiation — Question 10

Question 10 - Solution

The real domain is $x>-\tfrac12,\ x\ne3$. Where the function is nonzero, use absolute values in logarithmic differentiation:

$$\ln|y|=5\ln(x^2+1)+\tfrac13\ln|x^2-4|-\tfrac12\ln(2x+1)-2\ln|x-3|$$

Differentiating gives

$$\boxed{y'=y\left(\frac{10x}{x^2+1}+\frac{2x}{3(x^2-4)}-\frac1{2x+1}-\frac2{x-3}\right).}$$

This formula holds for $x>-\tfrac12,\ x\ne2,3$. At $x=2$, $y=0$, but there is no finite derivative because of the cube-root factor.