example 2. determine the derivatives of the following functions\n(a) (f(x)=ln x^{3})\n(b) (f(x)=lnleft(\frac{…

example 2. determine the derivatives of the following functions\n(a) (f(x)=ln x^{3})\n(b) (f(x)=lnleft(\frac{1}{x}\right))\n(c) (f(x)=lnleft(\frac{x}{x^{2}+1}\right))\n(d) (y = ln((x^{2}+2x)(sin x))^{2})\n(e) (f(x)=e^{2x}ln(cos x))
Answer
Explanation:
Step1: Recall derivative rule for $\ln(u)$
The derivative of $\ln(u)$ with respect to $x$ is $\frac{u'}{u}$ by the chain - rule, where $u'$ is the derivative of $u$ with respect to $x$. Also, recall the product rule $(uv)' = u'v+uv'$ for two functions $u$ and $v$.
Step2: Find derivative of $f(x)=\ln x^{3}$
Let $u = x^{3}$, then $u'=3x^{2}$. Using the chain - rule $\frac{d}{dx}\ln(u)=\frac{u'}{u}$, we have $\frac{d}{dx}\ln(x^{3})=\frac{3x^{2}}{x^{3}}=\frac{3}{x}$.
Step3: Find derivative of $f(x)=\ln(\frac{1}{x})$
Rewrite $f(x)=\ln(1)-\ln(x)=-\ln(x)$. The derivative of $-\ln(x)$ is $-\frac{1}{x}$.
Step4: Find derivative of $f(x)=\ln(\frac{x}{x^{2}+1})$
Using the property $\ln(\frac{a}{b})=\ln(a)-\ln(b)$, we have $f(x)=\ln(x)-\ln(x^{2}+1)$. The derivative of $\ln(x)$ is $\frac{1}{x}$ and the derivative of $\ln(x^{2}+1)$ is $\frac{2x}{x^{2}+1}$. So $f'(x)=\frac{1}{x}-\frac{2x}{x^{2}+1}=\frac{x^{2}+1 - 2x^{2}}{x(x^{2}+1)}=\frac{1 - x^{2}}{x(x^{2}+1)}$.
Step5: Find derivative of $y = [\ln((x^{2}+2x)\sin x)]^{2}$
Let $u=\ln((x^{2}+2x)\sin x)$. First, find the derivative of $(x^{2}+2x)\sin x$ using the product rule: $(x^{2}+2x)'\sin x+(x^{2}+2x)(\sin x)'=(2x + 2)\sin x+(x^{2}+2x)\cos x$. Then the derivative of $u$ is $\frac{(2x + 2)\sin x+(x^{2}+2x)\cos x}{(x^{2}+2x)\sin x}$. Now, using the chain - rule for $y = u^{2}$, $y'=2u\cdot u'=2\ln((x^{2}+2x)\sin x)\cdot\frac{(2x + 2)\sin x+(x^{2}+2x)\cos x}{(x^{2}+2x)\sin x}$.
Step6: Find derivative of $f(x)=e^{2x}\ln(\cos x)$
Using the product rule $(uv)' = u'v+uv'$ where $u = e^{2x}$ and $v=\ln(\cos x)$. The derivative of $u = e^{2x}$ is $2e^{2x}$, and the derivative of $v=\ln(\cos x)$ is $\frac{-\sin x}{\cos x}=-\tan x$. So $f'(x)=2e^{2x}\ln(\cos x)-e^{2x}\tan x=e^{2x}(2\ln(\cos x)-\tan x)$.
Answer:
(a) $\frac{3}{x}$ (b) $-\frac{1}{x}$ (c) $\frac{1 - x^{2}}{x(x^{2}+1)}$ (d) $2\ln((x^{2}+2x)\sin x)\cdot\frac{(2x + 2)\sin x+(x^{2}+2x)\cos x}{(x^{2}+2x)\sin x}$ (e) $e^{2x}(2\ln(\cos x)-\tan x)$