$$\begin{aligned} & \frac{d}{d x}\left(\frac{x^4+x^3+x^2+1}{x}\right)=\frac{d}{d x}\left(x^3+x^2+x+\frac{1}{x}\right) \\ & =\frac{d}{d x} x^3+\frac{d}{d x} x^2+\frac{d}{d x} x+\frac{d}{d x}\left(\frac{1}{x}\right) \\ & =3 x^2+2 x+1+\left(-\frac{1}{x^2}\right) \\ & =3 x^2+2 x+1-\frac{1}{x^2} \\ & =\frac{3 x^4+2 x^3+x^2-1}{x^2} \end{aligned} $$

$$\begin{aligned} \text { Let } \quad y & =\left(x+\frac{1}{x}\right)^3 \\ \therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(x+\frac{1}{x}\right)^3 & =3\left(x+\frac{1}{x}\right)^{3-1} \frac{d}{d x}\left(x+\frac{1}{x}\right) \quad \text{[by chain rule]}\\ & =3\left(x+\frac{1}{x}\right)^2\left(1-\frac{1}{x^2}\right) \\ & =3 x^2-\frac{3}{x^2}-\frac{3}{x^4}+3 \end{aligned}$$

$$\begin{aligned} &\text { Let } \quad y=(3 x+5)(1+\tan x)\\ &\therefore \quad \frac{d y}{d x}=\frac{d}{d x}[(3 x+5)(1+\tan x)] \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & =(3 x+5) \frac{d}{d x}(1+\tan x)+(1+\tan x) \frac{d}{d x}(3 x+5) \quad \text { [by product rule]}\\ & =(3 x+5)\left(\sec ^2 x\right)+(1+\tan x) \cdot 3 \\ & =(3 x+5) \sec ^2 x+3(1+\tan x) \\ & =3 x \sec ^2 x+5 \sec ^2 x+3 \tan x+3 \end{aligned}\\ \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text{Let}\quad y & =(\sec x-1)(\sec x+1) \\ y & =\left(\sec ^2-1\right) \quad\left[\because(a+b)(a-b)=a^2-b^2\right] \\ & =\tan ^2 x \\ \therefore \quad \frac{d y}{d x} & =2 \tan x \cdot \frac{d}{d x} \tan x \\ & =2 \tan x \cdot \sec ^2 x \quad \text { [by chain rule] } \end{aligned}\\ \end{aligned}$$

Let $\quad y=\frac{3 x+4}{5 x^2-7 x+9}$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{\left(5 x^2-7 x+9\right) \frac{d}{d x}(3 x+4)-(3 x+4) \frac{d}{d x}\left(5 x^2-7 x+9\right)}{\left(5 x^2-7 x+9\right)^2} \quad \text { [by quotient rule] } \\ & =\frac{\left(5 x^2-7 x+9\right) \cdot 3-(3 x+4)(10 x-7)}{\left(5 x^2-7 x+9\right)^2} \\ & =\frac{15 x^2-21 x+27-30 x^2+21 x-40 x+28}{\left(5 x^2-7 x+9\right)^2} \\ & =\frac{-15 x^2-40 x+55}{\left(5 x^2-7 x+9\right)^2} \\ & =\frac{55-15 x^2-40 x}{\left(5 x^2-7 x+9\right)^2} \end{aligned}$$