Question

ii. $f(x)=\frac{\tan(2x + 1)}{2x+1}$

Explanation:

Step1: Recall quotient - rule for differentiation

If $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \tan(2x + 1)$ and $v=2x + 1$.

Step2: Differentiate $u=\tan(2x + 1)$

Using the chain - rule, if $y=\tan(u)$ and $u = 2x+1$, then $\frac{du}{dx}=\sec^{2}(2x + 1)\cdot2 = 2\sec^{2}(2x + 1)$.

Step3: Differentiate $v = 2x+1$

$\frac{dv}{dx}=2$.

Step4: Apply the quotient - rule

$f^\prime(x)=\frac{2\sec^{2}(2x + 1)\cdot(2x + 1)-\tan(2x + 1)\cdot2}{(2x + 1)^{2}}=\frac{2(2x + 1)\sec^{2}(2x + 1)-2\tan(2x + 1)}{(2x + 1)^{2}}$.

Answer:

$\frac{2(2x + 1)\sec^{2}(2x + 1)-2\tan(2x + 1)}{(2x + 1)^{2}}$