How to do logarithmic differentiation

Question

$$y=\sqrt{\frac{x-1}{x^8+1}}$$

I know that you should take the natural log of both sides, so knowing that $\sqrt x=x^{1/2}$ and that $\ln(A/B)=\ln A-\ln B$ I got:

$$\ln y = \frac{1}{2} (\ln(x-1))-(\ln(x^8+1))$$

After taking the derivative of both sides I got:

$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} (1/(x-1))-(1/(x^8+1))$$

After multiplying the first term on the right by $1/2$ and then multiplying both sides by $y$ I got:

$$\frac{dy}{dx} = y((1/(2x-2))-(1/(x^8+1)))$$

I replaced $y$ with the original function, but this answer was wrong. I don't know where I went wrong, can anyone please help?

Answer 1

Full solution, mostly because I need LaTex speed practice :P

$$ y= \sqrt \frac {x-1}{x^8+1} \implies \ln y = \frac 12 \ln \frac {x-1}{x^8+1}$$ $$ 2 \ln y = \ln (x-1) - \ln (x^8+1) $$ Differentiate: $$ \frac {2y'}{y} = \frac {1}{x-1} - \frac {8x^7}{x^8+1}$$ $$y' = \frac 12 (\frac {1}{x-1} - \frac {8x^7}{x^8+1})\sqrt \frac {x-1}{x^8+1} \ \text{for} \ x \ge 1$$

Answer 2

we have $y=(x-1)^\frac{1}{2}(x^8+1)^\frac{-1}{2}$

thus

$\frac{dy}{y}=\frac{1}{2}\frac{dx}{x-1}-\frac{1}{2}\frac{8x^7dx}{x^8+1}$.

then simplify and factor out by $dx$.