example 2: determine the derivatives of the following functions:\n(a) $f(x)=ln x^{2}$\n(b)…

Explanation:

Step1: Recall log - property and chain - rule for (a)

First, use the property $\ln x^{2}=2\ln x$. The derivative of $\ln u$ with respect to $x$ is $\frac{u'}{u}$ by the chain - rule. For $y = 2\ln x$, and $u=x$, $u' = 1$. So $y'=2\times\frac{1}{x}=\frac{2}{x}$.

Step2: Simplify and differentiate (b)

Use the property $\ln(\frac{1}{x})=-\ln x$. The derivative of $-\ln x$: if $y =-\ln x$, and $u = x$, $u'=1$, then $y'=-\frac{1}{x}$.

Step3: Apply quotient - rule and chain - rule for (c)

First, if $y=\ln(\frac{x}{x^{2}+1})$, by the chain - rule $y'=\frac{(x^{2}+1)}{x}\times\frac{(x^{2}+1)\times1 - x\times2x}{(x^{2}+1)^{2}}$. Simplify the expression: [ \begin{align*} y'&=\frac{(x^{2}+1)}{x}\times\frac{x^{2}+1 - 2x^{2}}{(x^{2}+1)^{2}}\ &=\frac{x^{2}+1 - 2x^{2}}{x(x^{2}+1)}\ &=\frac{1 - x^{2}}{x(x^{2}+1)} \end{align*} ]

Answer:

(a) $f'(x)=\frac{2}{x}$ (b) $f'(x)=-\frac{1}{x}$ (c) $f'(x)=\frac{1 - x^{2}}{x(x^{2}+1)}$