$\text{a)}$  

$f\left(x\right)=\frac{3x-2}{x^2+2x+3}$  

Zał:

$x^2+2x+3\ne 0$  

$\Delta =2^2-4\cdot 1\cdot 3<0$  

$D_f=\mathbb{R}$  

Zatem:   

$f{}^{\prime }\left(x\right)=\frac{\left(3x-2\right){}^{\prime }\left(x^2+2x+3\right)-\left(3x-2\right)\left(x^2+2x+3\right){}^{\prime }}{{\left(x^2+2x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{3\left(x^2+2x+3\right)-\left(3x-2\right)\left(2x+2\right)}{{\left(x^2+2x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{3x^2+6x+9-6x^2-6x+4x+4}{{\left(x^2+2x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-3x^2+4x+13}{{\left(x^2+2x+3\right)}^2}$  

Zał:

${\left(x^2+2x+3\right)}^2\ne 0$  

$x^2+2x+3\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}$  

---

$\text{b)}$  

$f\left(x\right)=\frac{x^2-2}{x^2+3}$  

Zał:

$x^2+3\ne 0$  

$x^2\ne -3$  

$x\in \mathbb{R}$  

$D_f=\mathbb{R}$  

Zatem:

$f{}^{\prime }\left(x\right)=\frac{\left(x^2-2\right){}^{\prime }\left(x^2+3\right)-\left(x^2-2\right)\left(x^2+3\right){}^{\prime }}{{\left(x^2+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x\left(x^2+3\right)-\left(x^2-2\right)\cdot 2x}{{\left(x^2+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x^3+6x-2x^3+4x}{{\left(x^2+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{10x}{{\left(x^2+3\right)}^2}$  

Zał:

${\left(x^2+3\right)}^2\ne 0$  

$x^2+3\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}$  

---

$\text{c)}$  

$f\left(x\right)=\frac{x^2-x}{2x^2+x+3}$  

Zał:

$2x^2+x+3\ne 0$  

$\Delta =1^2-4\cdot 2\cdot 3<0$  

$D_f=\mathbb{R}$  

Zatem:

$f{}^{\prime }\left(x\right)=\frac{\left(x^2-x\right){}^{\prime }\left(2x^2+x+3\right)-\left(x^2-x\right)\left(2x^2+x+3\right){}^{\prime }}{{\left(2x^2+x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{\left(2x-1\right)\left(2x^2+x+3\right)-\left(x^2-x\right)\left(4x+1\right)}{{\left(2x^2+x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{4x^3+2x^2+6x-2x^2-x-3-4x^3-x^2+4x^2+x}{{\left(2x^2+x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{3x^2+6x-3}{{\left(2x^2+x+3\right)}^2}$  

Zał:

${\left(2x^2+x+3\right)}^2\ne 0$  

$2x^2+x+3\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}$  

---

$\text{d)}$  

$f\left(x\right)=\frac{x^2-3x-2}{x^2+3x+3}$  

Zał:

$x^2+3x+3\ne 0$  

$\Delta =3^2-4\cdot 1\cdot 3<0$  

$x\in \mathbb{R}$  

$D_f=\mathbb{R}$  

Zatem:

$f{}^{\prime }\left(x\right)=\frac{\left(x^2-3x-2\right){}^{\prime }\left(x^2+3x+3\right)-\left(x^2-3x-2\right)\left(x^2+3x+3\right){}^{\prime }}{{\left(x^2+3x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{\left(2x-3\right)\left(x^2+3x+3\right)-\left(x^2-3x-2\right)\left(2x+3\right)}{{\left(x^2+3x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x^3+6x^2+6x-3x^2-9x-9-\left(2x^3+3x^2-6x^2-9x-4x-6\right)}{{\left(x^2+3x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x^3+3x^2-3x-9-2x^3+3x^2+13x+6}{{\left(x^2+3x+3\right)}^2}$   

$f{}^{\prime }\left(x\right)=\frac{6x^2+10x-3}{{\left(x^2+3x+3\right)}^2}$  

Zał:

${\left(x^2+3x+3\right)}^2\ne 0$  

$x^2+3x+3\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}$  

---

$\text{e)}$  

$f\left(x\right)=\frac{x^2+1}{x^2-4}$  

Zał:

$x^2-4\ne 0$  

$\left(x-2\right)\left(x+2\right)\ne 0$  

$x\ne 2,x\ne -2$  

$D_f=\mathbb{R}-\left\{-2,2\right\}$  

Zatem:

$f{}^{\prime }\left(x\right)=\frac{\left(x^2+1\right){}^{\prime }\left(x^2-4\right)-\left(x^2+1\right)\left(x^2-4\right){}^{\prime }}{{\left(x^2-4\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x\left(x^2-4\right)-\left(x^2+1\right)\cdot 2x}{{\left(x^2-4\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{2x^3-8x-2x^3-2x}{{\left(x^2-4\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-10x}{{\left(x^2-4\right)}^2}$  

Zał:

${\left(x^2-4\right)}^2\ne 0$  

$x^2-4\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}-\left\{-2,2\right\}$  

---

$\text{f)}$  

$f\left(x\right)=\frac{2x^2}{{\left(2-x\right)}^2}$  

Zał:

${\left(2-x\right)}^2\ne 0$  

$2-x\ne 0$  

$2\ne x$  

$D_f=\mathbb{R}-\left\{2\right\}$  

Zatem:

$f{}^{\prime }\left(x\right)=\frac{\left(2x^2\right){}^{\prime }{\left(2-x\right)}^2-2x^2\left({\left(2-x\right)}^2\right){}^{\prime }}{{\left(2-x\right)}^4}$  

$f{}^{\prime }\left(x\right)=\frac{4x{\left(2-x\right)}^2-2x^2\left(4-4x+x^2\right){}^{\prime }}{{\left(2-x\right)}^4}$  

$f{}^{\prime }\left(x\right)=\frac{4x\left(4-4x+x^2\right)-2x^2\left(-4+2x\right)}{{\left(2-x\right)}^4}$  

$f{}^{\prime }\left(x\right)=\frac{16x-16x^2+4x^3+8x^2-4x^3}{{\left(2-x\right)}^4}$  

$f{}^{\prime }\left(x\right)=\frac{-8x^2+16x}{{\left(2-x\right)}^4}$  

$f{}^{\prime }\left(x\right)=\frac{8x\left(2-x\right)}{{\left(2-x\right)}^4}$  

$f{}^{\prime }\left(x\right)=\frac{8x}{{\left(2-x\right)}^3}$  

Zał:

${\left(2-x\right)}^4\ne 0$  

$2-x\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}-\left\{2\right\}$  

---

$\text{g)}$  

$f\left(x\right)=\frac{x^3}{2\left(4-x^2\right)}$  

Zał:

$2\left(4-x^2\right)\ne 0$  

$4-x^2\ne 0$  

$\left(2-x\right)\left(2+x\right)\ne 0$  

$x\ne 2,x\ne -2$  

$D_f=\mathbb{R}-\left\{-2,2\right\}$  

Zatem:

$f{}^{\prime }\left(x\right)=\frac{\left(x^3\right){}^{\prime }\left(2\left(4-x^2\right)\right)-x^3\left(2\left(4-x^2\right)\right){}^{\prime }}{{\left(2\left(4-x^2\right)\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{3x^2\left(8-2x^2\right)-x^3\left(8-2x^2\right){}^{\prime }}{4\cdot {\left(4-x^2\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{24x^2-6x^4-x^3\cdot \left(-4x\right)}{4\cdot {\left(4-x^2\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{24x^2-6x^4+4x^4}{4\cdot {\left(4-x^2\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{24x^2-2x^4}{4{\left(4-x^2\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-2x^2\left(-12+x^2\right)}{4{\left(4-x^2\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-x^2\left(x^2-12\right)}{2{\left(4-x^2\right)}^2}$  

 

Zał:

${\left(4-x^2\right)}^2\ne 0$  

$4-x^2\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}-\left\{-2,2\right\}$  

---

$\text{h)}$  

$f\left(x\right)=\frac{x-1}{x^2-7x+10}$  

Zał:

$x^2-7x+10\ne 0$  

$\Delta ={\left(-7\right)}^2-4\cdot 1\cdot 10=49-40=9$  

$x_1=\frac{7-3}{2}=\frac{4}{2}=2$  

$x_2=\frac{7+3}{2}=\frac{10}{2}=5$  

$D_f=\mathbb{R}-\left\{2,5\right\}$  

Zatem:

$f{}^{\prime }\left(x\right)=\frac{\left(x-1\right){}^{\prime }\left(x^2-7x+10\right)-\left(x-1\right)\left(x^2-7x+10\right){}^{\prime }}{{\left(x^2-7x+10\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{1\left(x^2-7x+10\right)-\left(x-1\right)\left(2x-7\right)}{{\left(x^2-7x+10\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{x^2-7x+10-2x^2+7x+2x0-7}{{\left(x^2-7x+10\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-x^2+2x+3}{{\left(x^2-7x+10\right)}^2}$  

Zał:

${\left(x^2-7x+10\right)}^2\ne 0$  

$x^2-7x+10\ne 0$  

$D_{f{}^{\prime }}=\mathbb{R}-\left\{2,5\right\}$