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$$G(y) = \ln \dfrac{(5y+1)^2}{\sqrt{y^2 + 1}}$$

Can we break this up like this: $$2 \ln(5y+1)\over \frac 12\ln(y^2 + 1)$$

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$\ln\left(\dfrac{f(y)}{g(y)} \right) \neq \dfrac{\ln(f(y))}{\ln(g(y))}$. But make use of the fact that $$\ln\left(\dfrac{f(y)}{g(y)} \right) = \ln(f(y)) - \ln(g(y))$$ and $$\dfrac{d \ln(h(y))}{dy} = \dfrac{h'(y)}{h(y)}$$

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$$G(y) = \ln\left(\dfrac{(5y+1)^2}{\sqrt{y^2+1}} \right) = \ln((5y+1)^2) - \ln(\sqrt{y^2+1}) = 2 \ln(5y+1) - \dfrac12 \ln(y^2+1)$$Hence, \begin{align}\dfrac{dG(y)}{dy} & = 2 \times \dfrac1{5y+1} \times 5 - \dfrac12 \times \dfrac1{y^2+1} \times (2y) \\& = \dfrac{10}{5y+1} - \dfrac{y}{y^2+1}\\ & = \dfrac{10y^2 + 10 - 5y^2 - y}{(5y+1)(y^2+1)}\\ & = \dfrac{5y^2 - y +10}{(5y+1)(y^2+1)} \end{align}

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So I can write it as: $2ln(5y+1) - \frac 12ln(y^2 + 1)$? –  dsta Oct 20 '12 at 21:11
    
@dsta Yes. You should write it like that and proceed with the differentiation. –  user17762 Oct 20 '12 at 21:13
    
@dsta Yes. It should be $$2 \times \dfrac5{5y+1} - \dfrac12 \dfrac{2y}{y^2+1}$$ –  user17762 Oct 20 '12 at 21:17
    
So in the end I have: $10\over 5y+1$ - $2y\over 2y^2+1$= $2\over y+1$ - $1\over y+1$= $ 1\over y+1$? –  dsta Oct 20 '12 at 21:20
    
@dsta No your manipulation is incorrect. I have now added the complete solution. –  user17762 Oct 20 '12 at 21:28
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No: is $\ln\frac{a}b$ equal to $\frac{\ln a}{\ln b}$, or is it equal to some other combination of $\ln a$ and $\ln b$? (You did handle the exponents correctly, though.)

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