Question

2021: q3. a. find the derivative of the following. you do not need to simplify your answer.
i. f(x)=6ln(x)
ii. f(x)=\frac{\tan(2x + 1)}{2x + 1}

Explanation:

Step1: Recall derivative formula for $a\ln(x)$

The derivative of $\ln(x)$ is $\frac{1}{x}$, and by the constant - multiple rule, if $y = a\ln(x)$ where $a$ is a constant, then $y^\prime=a\cdot\frac{1}{x}$. For $f(x)=6\ln(x)$, we have:
$f^\prime(x)=6\cdot\frac{1}{x}=\frac{6}{x}$

Step2: Recall the quotient rule for $f(x)=\frac{u(x)}{v(x)}$

The quotient rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2}$. Here, $u(x)=\tan(2x + 1)$ and $v(x)=2x + 1$. First, find $u^\prime(x)$ and $v^\prime(x)$. The derivative of $\tan(u)$ with respect to $x$ is $\sec^{2}(u)\cdot u^\prime$ (by the chain - rule). For $u = 2x+1$, $u^\prime=2$, so $u^\prime(x)=2\sec^{2}(2x + 1)$, and $v^\prime(x)=2$. Then:
$f^\prime(x)=\frac{2\sec^{2}(2x + 1)\cdot(2x + 1)-\tan(2x + 1)\cdot2}{(2x + 1)^2}$

Answer:

i. $f^\prime(x)=\frac{6}{x}$
ii. $f^\prime(x)=\frac{2(2x + 1)\sec^{2}(2x + 1)-2\tan(2x + 1)}{(2x + 1)^2}$