Derived gates are the gates which are derived from the fundamental logical gates. Ex. NAND, NOR, XOR, XNOR.

A + (B + C) = ( A + B) + C
A. (B. C) = (A . B) .C

Boolean algebra is a mathematical discipline that is used for designing digital circuits in a digital computer. It describes the relation between inputs and outputs of a digital circuit.

Step 1: Convert given Decimal number into Binary
Step 2: Check if the binary number contains 8 bits if less add 0 at the left most bit, to make it as 8 bits.
Step 3: Invert all bits (i.e. Change 1 as 0 and 0 as 1).

The term data comes from the word datum, which means a raw fact. The data is a fact about people. places or some objects.
Example : Rajesh 16, XI

(3674)₈ in binary From
      3     6      7     4
      011  110   111  100
(3674)₈ = (011110111100)₂
(3674)₈ in decimal Form 
(3674)₈ = 3 x 8³ + x 8² + 7 x 8¹ + 4 x 8⁰
= 1536 + 384 + 56 + 4 = (1980)₁₀
Thus (3674)₈ \(\rightarrow \) (0111110111100)₂ = (1980)₁₀

The logic gates like NAND, NOR, XOR and XNOR are derived gates which are derived from the fundamental gates AND, OR NOT.
(1) NAND gate:
The NAND is the combination of NOT and AND. The NAND is generated by inverting the output of an AND operation.
Output: y = \(\overline { \left( A.B \right) } \)

The truth table for NAND gate is

(ii) NOR gate:
The NOR is the combination of NOT and OR. The NOR is generated by inverting the output of an OR operation.
Logic function: y = \(\overline { \left( A+B \right) } \)

the truth table for NAND gate is

(iii) XOR gate:
The XOR (exclusive - OR) gate acts in the same way as the logical either /or.

the truth table 

(iv) XNOR gate:
XNOR (exclusive - NOR) gate is a combination of XOR gate followed by an inverter.
Logic function: y= \(\overline { A \ B } \)

the truth table 

Boolean function:
\(\mathrm{C} =\overline{\mathrm{A} \oplus \mathrm{B}} \)
\(=\overline{{\overline{\mathrm{A}}} \cdot \mathrm{B}+\mathrm{A} \cdot \overline{\mathrm{B}}} \)
\(=\overline{\overline{\mathrm{A}} \cdot \mathrm{B}} \cdot \overline{\mathrm{A} \cdot \overline{\mathrm{B}}} \)
\(=(\overline{\mathrm{A}}+\overline{\mathrm{B}})(\overline{\mathrm{A}}+\overline{\bar{B}}) \)
\(=(\mathrm{A}+\overline{\mathrm{B}})(\overline{\mathrm{A}}+\mathrm{B}) \)
\(=\mathrm{A} \overline{\mathrm{A}}+\overline{\mathrm{AB}}+\mathrm{AB}+\mathrm{B} \overline{\mathrm{B}}  \quad \mathrm{A \overline {A }}=\mathrm{O}\)
\(\mathrm{C} =\overline{\mathrm{A}} \overline{\mathrm{B}}+\mathrm{AB}\)

1101010₂ + 101101₂
1101010
+101101
________
10010111
________
1101010₂ + 101101₂  = 10010111₂

The method of repeated multiplication by 2 has to be used to convert such kinds of decimal fractions.
The steps involved in the method of repeated multiplication by 2.
Step 1: Multiply the decimal fraction by 2 and note. the integer part. The integer part is either 0 or 1.
Step 2: Discard the integer part of the previous product. Multiply the fractional part of the previous product by 2. Repeat Step 1 until the same fraction repeats or terminates (0).
Step 3: The resulting integer part forms a sequence of 0's and 1's that become the binary equivalent of a decimal fractions.
Step 4: The final answer is to be written from first. integer part obtained till the last integer part is obtained.

The radix refers to the base of a number system: the total number of possible digits.
The decimal number system that we all use is base ten, as it has ten distinct digits (0,1,2,3,4,5,6,7,8,9). Eg: Binary system- Radix 2

Step 1: Find 1's complement (Replace 1 by 0 and 0 by 1
Step 2: Add 1 to the 1's complement

(365)₈ = 3 x 8² x 6 x 8¹ + 5 x 8⁰
= 192 + 48 + 5 = (245)₁₀
(366)₈ = 3 x 8² + 6 x 8¹ + 6 x 8⁰
= 192 + 42 + 6 = (246)₁₀
(367)₈ = 3 x 8² + 6 x 8¹ + 7 x 8⁰
= 192 + 42 + 6 = (247)₁₀
In decimal form, the given sequence will be 245, 246, 247, 248, 249, 250.
To complete the sequence of Octal number, we need to find the octal equal of 248, 249 and 250

(COFFEE)₁₆
\(\overset { C }{ \underset { 1100 }{ \downarrow } } \)\(\overset { O }{ \underset { 0000 }{ \downarrow } } \)\(\overset { F }{ \underset { 1111 }{ \downarrow } } \)\(\overset { F }{ \underset { 1111 }{ \downarrow } } \)\(\overset { E }{ \underset { 1110 }{ \downarrow } } \)\(\overset { E }{ \underset { 1110 }{ \downarrow } } \)
(COFFEE)₁₆  = (110000001111111111101110)₂

De Morgan's
\(\bar { A+B } =\bar { A } .\bar { B } \)
\(\left( \bar { A.B } \right) =\bar { A } +\bar { B } \)