Question

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

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1. -/1 points

differentiate the function.
$f(s)=ln(ln(5s))$
$f(s)=$

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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:

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