Logarithmic Differentiation — Question 8

Question 8 - Solution

Put $A=x^2+3x+1$. The real domain is $|x|\geq1$, $x\ne-5/2$.

For $A\ne0$ and $|x|>1$, logarithmic differentiation gives

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

$$\frac{y'}y=\frac{4(2x+3)}A+\frac{x}{x^2-1}-\frac6{2x+5}.$$

Cancel before evaluating at zeros of $A$:

$$\boxed{y'=\frac{A^3\big[4(2x+3)(x^2-1)(2x+5)+Ax(2x+5)-6A(x^2-1)\big]}{\sqrt{x^2-1}(2x+5)^4}.}$$

This formula holds for $|x|>1$, $x\ne-5/2$, including $x=(-3-\sqrt5)/2$, where $y'=0$. The other root of $A$ is outside the domain.

There is no finite derivative at $x=\pm1$.