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Title: Differentiation
Description: Differentiation-Rules Exercise Questions and Answers

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1)
Derivative the basic rule:
Where the Function is 𝑦 = π‘₯ n
The Derivative 𝑑𝑦 /𝑑π‘₯ = n x π‘₯ (n-1)

Example:Function
𝑦 =π‘₯ 2

Derivatives
𝑑𝑦 / 𝑑π‘₯ = 2 x π‘₯ (2-1) = 2π‘₯

Questions and Answers
I
...


𝑦 =π‘₯ -2

𝑑𝑦 / 𝑑π‘₯ = -2 x π‘₯ (-2-1) = -2 π‘₯ -3

III
...


𝑦 =8π‘₯ 4

𝑑𝑦 / 𝑑π‘₯ =4 x 8π‘₯ (4-1) = 32π‘₯ 3

II
...


𝑦 =1/2π‘₯ 9

𝑑𝑦 / 𝑑π‘₯ =9 x 1/2π‘₯ (9-1) = 4
...


𝑦 = 6π‘₯ 4+27

𝑑𝑦 / 𝑑π‘₯ = 4 x 6π‘₯ (4-1) = 24π‘₯ 3

II
...


𝑦= 1/6π‘₯ 2 - 12π‘₯ 5

𝑑𝑦 / 𝑑π‘₯ = 2 x 1/6π‘₯ (2-1) - 5 x 12π‘₯ (5-1)
= 1/3π‘₯ - 60π‘₯ 4

II
...
𝑑𝑛/ 𝑑π‘₯
𝑛2
= π‘₯ 2(21π‘₯ 2+1) – (7π‘₯ 3+π‘₯) 2π‘₯
(π‘₯ 2)2
= 21π‘₯ 4 + π‘₯2 -14π‘₯ 4 - 2π‘₯ 2
π‘₯4
= 7π‘₯ 4 - π‘₯ 2
π‘₯4

𝑉𝐼𝐼) 𝑦 = 4π‘₯ 2 + 3

π’Ž

2π‘₯ -1

𝒏

𝑑𝑦 / 𝑑π‘₯ = 𝑛 x π‘‘π‘š/ 𝑑π‘₯ – π‘š x 𝑑𝑛/ 𝑑π‘₯
(2π‘₯ -1)2
= (2π‘₯-1)8π‘₯ – (4π‘₯ 2+3)2
4π‘₯ 2 - 4π‘₯ +1
= 16π‘₯ 2 - 8π‘₯ - 8π‘₯ 2 - 6
4π‘₯ 2 - 4π‘₯+1
= 8π‘₯ 2-8π‘₯-6
4π‘₯ 2-4π‘₯+1

𝑉𝐼𝐼𝐼) 𝑦 = 4 - 5π‘₯
(2π‘₯-1) (3π‘₯+2)

π’Ž
𝒏

𝑑𝑦 / 𝑑π‘₯ = n x π‘‘π‘š/ 𝑑π‘₯ – m x 𝑑𝑛/ 𝑑π‘₯
𝑛2
= [(2π‘₯-1) (3π‘₯+2)]-5 – [(4-5π‘₯) (12π‘₯+1)]
[6π‘₯ 2 + π‘₯ -2]2
= (6π‘₯ 2+4π‘₯-3π‘₯-2)-5 - (48π‘₯ + 4 - 60π‘₯ 2 - 5π‘₯)
36π‘₯ 4 + 12π‘₯3 - 23π‘₯ 2 + 4

= (-30 π‘₯2 - 20π‘₯ + 15π‘₯+10) (-48π‘₯ – 4 + 60π‘₯ 2 + 5π‘₯)

36π‘₯ 4 + 12π‘₯3 - 23π‘₯ 2 + 4
= -30π‘₯ 2 - 5π‘₯ +10 - 43π‘₯ + 60π‘₯ 2 - 4
36π‘₯ 4 + 12π‘₯3 - 23π‘₯ 2 + 4
= 30π‘₯ 2 - 48π‘₯ + 6
36π‘₯ 4 + 12π‘₯3 - 23π‘₯ 2 + 4
7)
Derivation function of a function where π’š= (2𝒙+6)3 and the expression in the
brackets is a derivatization functions say that is π’Ž = (2𝒙+6) the code
expression can be written as π’š=π’Ž3
In that cases he rule for derivation is π’…π’š / 𝒅𝒙 = π’…π’š / π’…π’Ž x π’…π’Ž/ 𝒅𝒙
Which is known as the chain rule
...
π‘‘π‘š/ 𝑑π‘₯
= 4π‘₯ x 1/π‘₯ + π‘₯ x 4
= 4π‘₯ (1/π‘₯+1)
g)
π’Ž
𝒏
𝑦 = (1+ π‘₯2) 𝑒ax
𝑑𝑦 / 𝑑π‘₯ = (1+π‘₯ 2) π‘Žπ‘’ax + 𝑒ax x 2x
𝑒ax (1+π‘₯ 2)x a +2π‘₯

h)
π’Ž 𝒏
𝑦 = eπ‘₯ In π‘₯
𝑑𝑦 / 𝑑π‘₯ = eπ‘₯ x 1/π‘₯ + π‘₯ x 𝑒x
𝑒x (1/π‘₯+π‘₯)


Title: Differentiation
Description: Differentiation-Rules Exercise Questions and Answers