differentiate the function.\n$g(y)=ln\\left(\\frac{(2y + 1)^{4}}{\\sqrt{y^{2}+1}}\\right)$\n$g(y)=$\n\nresour…

Explanation:

Step1: Use log - properties to simplify (G(y))

We know that (\ln(\frac{a}{b})=\ln a-\ln b) and (\ln(a^n) = n\ln a). So, (G(y)=\ln((2y + 1)^4)-\ln((y^{2}+1)^{\frac{1}{2}})=4\ln(2y + 1)-\frac{1}{2}\ln(y^{2}+1)).

Step2: Differentiate term - by - term using the chain rule

The derivative of (\ln(u)) with respect to (y) is (\frac{u'}{u}). For (u = 2y+1), (u'=2), and for (u = y^{2}+1), (u' = 2y). The derivative of (4\ln(2y + 1)) is (4\times\frac{2}{2y + 1}=\frac{8}{2y+1}). The derivative of (-\frac{1}{2}\ln(y^{2}+1)) is (-\frac{1}{2}\times\frac{2y}{y^{2}+1}=-\frac{y}{y^{2}+1}). So, (G'(y)=\frac{8}{2y + 1}-\frac{y}{y^{2}+1}).

Step3: Differentiate (F(s)=\ln(\ln(5s))) using the chain rule

Let (u=\ln(5s)), then (F(s)=\ln(u)). The derivative of (\ln(u)) with respect to (s) is (\frac{u'}{u}). First, find the derivative of (u = \ln(5s)). Since (\ln(5s)=\ln(5)+\ln(s)) and the derivative of (\ln(5)) (a constant) is (0) and the derivative of (\ln(s)) is (\frac{1}{s}), (u'=\frac{1}{s}). So, (F'(s)=\frac{\frac{1}{s}}{\ln(5s)}=\frac{1}{s\ln(5s)}).

Answer:

(G'(y)=\frac{8}{2y + 1}-\frac{y}{y^{2}+1}), (F'(s)=\frac{1}{s\ln(5s)})