find the indicated derivative and simplify.\n$y$ for $y = \\frac{\\log_{5}x}{4 + x^{5}}$

Explanation:

Step1: Recall quotient - rule and derivative of log

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Also, the derivative of $\log_{a}x=\frac{1}{x\ln a}$. Here, $u = \log_{5}x=\frac{\ln x}{\ln 5}$, so $u'=\frac{1}{x\ln 5}$, and $v = 4 + x^{5}$, so $v'=5x^{4}$.

Step2: Apply the quotient - rule

$y'=\frac{\frac{1}{x\ln 5}(4 + x^{5})-\log_{5}x\cdot5x^{4}}{(4 + x^{5})^{2}}$.

Step3: Simplify the expression

[ \begin{align*} y'&=\frac{\frac{4}{x\ln 5}+\frac{x^{5}}{x\ln 5}- 5x^{4}\log_{5}x}{(4 + x^{5})^{2}}\ &=\frac{\frac{4}{x\ln 5}+\frac{x^{4}}{\ln 5}-5x^{4}\log_{5}x}{(4 + x^{5})^{2}}\ &=\frac{\frac{4 + x^{4}}{x\ln 5}-5x^{4}\log_{5}x}{(4 + x^{5})^{2}}\ &=\frac{4 + x^{4}-5x^{5}\log_{5}x}{x\ln 5(4 + x^{5})^{2}} \end{align*} ]

Answer:

$\frac{4 + x^{4}-5x^{5}\log_{5}x}{x\ln 5(4 + x^{5})^{2}}$