defination and rules of derivatives and application

1. Derivatives
WEIGHTAGE : 17 MARKS

2. USING DEFINITION OBTAIN DERIVATIVE
• Find derivative of y = 𝒙
Here f(x)= 𝒙
Therefore f(x+h)= 𝒙 + 𝒉
Now according to definition of derivative
• f’(x)= 𝐥𝐢𝐦
𝒉→𝟎
𝒇 𝒙+𝒉 −𝒇(𝒙)
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
𝒙+𝒉− 𝒙
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
𝒙+𝒉− 𝒙
𝒉
×
𝒙+𝒉+ 𝒙
𝒙+𝒉+ 𝒙
• = 𝐥𝐢𝐦
𝒉→𝟎
𝒙+𝒉
𝟐
− 𝒙
𝟐
𝒉( 𝒙+𝒉+ 𝒙)
= lim
ℎ→0
𝑥+ℎ−𝑥
ℎ ( 𝑥+ℎ+ 𝑥)
= lim
ℎ→0
ℎ
ℎ ( 𝑥+ℎ+ 𝑥)
= lim
ℎ→0
1
( 𝑥 + ℎ + 𝑥)
=
1
( 𝑥+0+ 𝑥)
=
1
( 𝑥 + 𝑥)
=
1
2 𝑥

3. DERIVATIVE OF CONSTANT ( C )
• Find derivative of y =c
Here f(x)=c
Therefore f(x+h)=c
Now according to definition of derivative
• f’(x)= 𝐥𝐢𝐦
𝒉→𝟎
𝒇 𝒙+𝒉 −𝒇(𝒙)
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
𝒄−𝒄
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
𝟎
𝒉
• =0

4. Using definition find derivatives
• Find derivative y=3𝑥2
− 2
Here f(x)= 3𝑥2
− 2
Therefore f(x+h)= 3 𝑥 + ℎ 2
− 2
=3(𝒙𝟐
+2xh+𝒉𝟐
)-2
= 3𝒙𝟐
+6xh+3𝒉𝟐
-2
Now according to definition of derivative
• f’(x)= 𝐥𝐢𝐦
𝒉→𝟎
𝒇 𝒙+𝒉 −𝒇(𝒙)
𝒉
=𝐥𝐢𝐦
𝒉→𝟎
3𝒙𝟐+6xh+3𝒉𝟐−2 −(3𝑥2 −2)
𝒉
=𝐥𝐢𝐦
𝒉→𝟎
3𝒙𝟐+6xh+3𝒉𝟐−2−3𝑥2+𝟐
𝒉
• =𝐥𝐢𝐦
𝒉→𝟎
6xh+3𝒉𝟐
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
𝒉(6x+3𝒉)
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
(𝟔𝒙 + 𝒉)
• = 𝟔𝒙 + 𝟎
• = 𝟔𝒙

5. Using definition find derivatives
• Find derivative y=4𝑥2
+ 5x − 2
Here f(x)= 4𝑥2
+ 5x − 2
Therefore f(x+h)= 4 𝑥 + ℎ 2
+ 5(x + h) − 2
=4(𝒙𝟐
+2xh+𝒉𝟐
) + 𝟓(𝐱 + 𝐡)-2
= 4𝒙𝟐
+8xh+4𝒉𝟐
+5x+5h-2
Now according to definition of derivative
• f’(x)= 𝐥𝐢𝐦
𝒉→𝟎
𝒇 𝒙+𝒉 −𝒇(𝒙)
𝒉
=𝐥𝐢𝐦
𝒉→𝟎
(4𝒙𝟐+8xh+4𝒉𝟐+5x+5h−2)−(4𝑥2+5x−2)
𝒉
=𝐥𝐢𝐦
𝒉→𝟎
4𝒙𝟐+8xh+4𝒉𝟐+5x+5h−2−4𝑥2−5x+2
𝒉
• =𝐥𝐢𝐦
𝒉→𝟎
8xh+4𝒉𝟐+5h
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
𝒉(8x+4h+5)
𝒉
• = 𝐥𝐢𝐦
𝒉→𝟎
(8x+4h+5)
• = 8𝒙 + 𝟒(𝟎) + 𝟓
• = 8𝒙+5

6. IMPORTANT FORMULAE FOR
DIFFERENTIATION
• If y = 𝑥𝑛
then
𝑑𝑦
𝑑𝑥
=n𝑥𝑛−1
• If y = x then
𝑑𝑦
𝑑𝑥
= 1
• If y=c then
𝑑𝑦
𝑑𝑥
=0
• If y=𝑒𝑥
then
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
• If y= 𝑎𝑥
then
𝑑𝑦
𝑑𝑥
= 𝑎𝑥
loge a
• If y= loge x then
𝑑𝑦
𝑑𝑥
=
1
𝑥

7. RULES OF DERIVATIVES
• If U and V are functions of x then
y=u+v then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢+𝑣)
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
+
𝑑𝑣
𝑑𝑥
y=u−v then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢−𝑣)
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
-
𝑑𝑣
𝑑𝑥
y=u×v then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢𝑣)
𝑑𝑥
= 𝑣
𝑑𝑢
𝑑𝑥
+ u
𝑑𝑣
𝑑𝑥
Y=uvw then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢𝑣𝑤)
𝑑𝑥
= 𝑣𝑤
𝑑𝑢
𝑑𝑥
+ 𝑢𝑤
𝑑𝑣
𝑑𝑥
+ 𝑢𝑣
𝑑𝑤
𝑑𝑥
Y=
𝑢
𝑣
then
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
Chain rule : If y is function of u and u is function of x then
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
×
𝑑𝑢
𝑑𝑥

8. If U and V are functions of x then (I)y=u+v then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢+𝑣)
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
+
𝑑𝑣
𝑑𝑥
(II)y=u−v then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢−𝑣)
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
-
𝑑𝑣
𝑑𝑥
Find
𝑑𝑦
𝑑𝑥
of the following functions using rule (i) and (ii)
(I) 3𝑥2
+2
SOLUTION :
I. Y= 3𝑥2
+2
𝑑𝑦
𝑑𝑥
=
𝑑3𝑥2
𝑑𝑥
+
𝑑2
𝑑𝑥
[∵ RULE (I)
APPLIED ]
= 3(2X)+0
=6X
(ii)4𝑥2
+5X+1
SOLUTION :
Y= 4𝑥2
+5X+1
𝑑𝑦
𝑑𝑥
=
𝑑4𝑥2
𝑑𝑥
+
𝑑5𝑥
𝑑𝑥
+
𝑑1
𝑑𝑥
= 4(2X)+5(1) +0
= 8X+5
(III) 5X-9
SOLUTION :
Y=5X-9
𝑑𝑦
𝑑𝑥
=
𝑑 5x
𝑑𝑥
-
𝑑9
𝑑𝑥
=5(1)-0
=5

9. Solve the following
(i) y=𝑥3
+ 𝑥2
+ 𝑥 − 2 (II) y=7𝑥7
+ 5𝑥4
− 20𝑥 +
37
• SOLUTION :
• y=𝑥3
+ 𝑥2
+ 𝑥 − 2
•
𝑑𝑦
𝑑𝑥
=
𝑑𝑥3
𝑑𝑥
+
𝑑 𝑥2
𝑑𝑥
+
𝑑𝑥
𝑑𝑥
-
𝑑2
𝑑𝑥
• =(3𝑥3−1
) +(2𝑥)+1-0
• = 3 𝑥2
+2𝑥+1
• SOLUTION:
• y= 7𝑥7
+ 5𝑥4
− 20𝑥 + 37
•
𝑑𝑦
𝑑𝑥
=
𝑑7𝑥7
𝑑𝑥
+
𝑑 5𝑥4
𝑑𝑥
-
𝑑20𝑥
𝑑𝑥
+
𝑑37
𝑑𝑥
• =7(7𝑥7−1
) +5(4𝑥4−1
)-
20+0
• = 49𝑥6
+20𝑥3
− 20

10. (I) y=u×v then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢𝑣)
𝑑𝑥
= 𝑣
𝑑𝑢
𝑑𝑥
+ u
𝑑𝑣
𝑑𝑥
II Y=uvw then
𝑑𝑦
𝑑𝑥
=
𝑑(𝑢𝑣𝑤)
𝑑𝑥
= 𝑣𝑤
𝑑𝑢
𝑑𝑥
+ 𝑢𝑤
𝑑𝑣
𝑑𝑥
+
𝑢𝑣
𝑑𝑤
𝑑𝑥
• Find the derivative w.r.t 𝑥
• Y= (3𝑥2
− 2)(𝑥2
+ 7)
• 𝑢 = 3𝑥2 − 2 𝑣 = (𝑥2 + 7)
•
𝑑𝑢
𝑑𝑥
=6𝑥
𝑑𝑣
𝑑𝑥
=2𝑥
•
𝑑𝑦
𝑑𝑥
= 𝑣
𝑑𝑢
𝑑𝑥
+ u
𝑑𝑣
𝑑𝑥
• =(𝑥2 + 7)(6𝑥) + 3𝑥2 − 2 (2𝑥)
• =(6𝑥3+42𝑥) +(6𝑥3-4𝑥)
• = 6𝑥3
+42𝑥+ 6𝑥3
-4𝑥
• = 12𝑥3 +38𝑥
• Find the derivative w.r.t 𝑥
• Y= (2𝑥2
+ 3)(3𝑥2
+ 2)(𝑥+7)
• 𝑢 = (2𝑥2 + 3) 𝑣 = 3𝑥2 + 2 𝑤 = (𝑥+7)
•
𝑑𝑢
𝑑𝑥
= 4𝑥 + 0
𝑑𝑣
𝑑𝑥
= 6𝑥 + 0
𝑑𝑤
𝑑𝑥
= 1+0
•
𝑑𝑦
𝑑𝑥
= 𝑣𝑤
𝑑𝑢
𝑑𝑥
+ 𝑢𝑤
𝑑𝑣
𝑑𝑥
+ 𝑢𝑣
𝑑𝑤
𝑑𝑥
=(3𝑥2 + 2)(𝑥+7) 4𝑥+ (2𝑥2 + 3) (𝑥+7) 6𝑥+ (2𝑥2 + 3)(3𝑥2 + 2)1
= (3𝑥2 + 2)(4𝑥2+28𝑥)+ (2𝑥2 + 3)(6𝑥2+42𝑥) +(2𝑥2 + 3)(3𝑥2 + 2)
= 4𝑥2
(3𝑥2
+ 2)+ 28𝑥 (3𝑥2
+ 2)+ 6𝑥2
(2𝑥2
+ 3)+42𝑥(2𝑥2
+ 3)+ 3𝑥2
(2𝑥2
+ 3) + 2(2𝑥2
+ 3)
=12𝑥4
+ 8𝑥2
+ 84𝑥3
+ 56𝑥 + 12𝑥4
+ 18𝑥2
+ 84𝑥3
+ 126𝑥 + 6𝑥4
+9𝑥2
+4𝑥2
+ 6
= 12𝑥4
+ 12𝑥4
+ 6𝑥4
+84𝑥3
+84𝑥3
+ 8𝑥2
+18𝑥2
+9𝑥2
+4𝑥2
+56𝑥 + 126𝑥+6
=30𝑥4
+ 168𝑥3
+ 39𝑥2
+ 182𝑥 + 6

11. NOW TRY YOUR SELF
• FIND DERIVATIVES WITH RESPECT TO 𝑥
• Y=(2𝑥 + 3) 5𝑥3 − 2
• Y=𝑥4
+ 𝑥5
+ 5𝑥6
+ 3𝑥2
+ 3𝑥3
− 9
• Y=3𝑥4 + 13𝑥3 − 9𝑥 + 15
• Y=(𝑥 + 1)(9𝑥 + 5)(𝑥2 − 9)
• Y= 3𝑥2 + 5𝑥 − 9
• Y=(𝑥2
+ 𝑥 − 1)(𝑥2
+ 𝑥 + 1)
• Y=(2X-1)(6X+1)
• Y=7𝑥2 + 123𝑥 − 100

12. Y=
𝑢
𝑣
then
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
• FIND THE DERIVATIVE OF 𝑦 =
1+𝑥
1+𝑥2
• HERE 𝑦 =
1+𝑥
1+𝑥2 =
𝑢
𝑣
• LET 𝑢 = (1 + 𝑥) 𝑣 = (1 + 𝑥2)
•
𝑑𝑢
𝑑𝑥
= 0 + 1
𝑑𝑣
𝑑𝑥
= 0 + 2𝑥
•
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
• =
(1+𝑥2) (1)−(1+𝑥)(2x)
(1+𝑥2)2
=
1 + 𝑥2
− 2𝑥 − 2𝑥2
(1 + 𝑥2)2
=
1−2𝑥−𝑥2
(1+𝑥2)2
• FIND DERIVATIVE 𝑦 =
𝑥2+𝑥+1
𝑥2−𝑥+1
• HERE 𝑦 =
𝑥2+𝑥+1
𝑥2−𝑥+1
=
𝑢
𝑣
• LET 𝑢 = (𝑥2
+ 𝑥 + 1) 𝑣 = (𝑥2
− 𝑥 +
1)
•
𝑑𝑢
𝑑𝑥
= 2𝑥 + 1
𝑑𝑣
𝑑𝑥
= 2𝑥 − 1
•
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
•
=
(𝑥2−𝑥+1) (2X+1)−(𝑥2+𝑥+1)(2x−1)
(1−𝑥+𝑥2)2
=
=

13. Find derivati𝑣𝑒𝑠 w.r.t. 𝑥
• Y=
3
2−5𝑥
• HERE 𝑦 =
3
2−5𝑥
=
𝑢
𝑣
• LET 𝑢 = 3 𝑣 = (2 − 5𝑥)
•
𝑑𝑢
𝑑𝑥
= 0
𝑑𝑣
𝑑𝑥
= −5
•
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
• =
(2−5𝑥)(0)−3(−5)
(2−5𝑥)2
=
0 + 15
(2 − 5𝑥)2
=
15
(2−5𝑥)2
• Y=1+
1
1+
1
𝑥
• Y=1+
1
1
1
+
1
𝑥
=1+
1
𝑥+1
𝑥
=1+
𝑥
𝑥+1
=
1
1
+
𝑥
𝑥+1
=
𝑥+1+𝑥
𝑥+1
=
2𝑥+1
𝑥+1
• NOW 𝑦 =
2𝑥+1
𝑥+1
• 𝑢 = 2𝑥 + 1 𝑣 =
•
𝑑𝑢
𝑑𝑥
= 2
𝑑𝑣
𝑑𝑥
= 1
•
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2 =
𝑥+1 2 −(2𝑥+1)(1)
(𝑥+1)2 =
2𝑥+2−2𝑥−1
(𝑥+1)2 =
1
(𝑥+1)2

14. Find derivative of
3
(𝑥+2)(𝑥+3)
• 𝑦 =
3
(𝑥+2)(𝑥+3)
• 𝑦 =
3
𝑥2+3𝑥+2𝑥+6
• ∴ 𝑦=
3
𝑥2+5𝑥+6
• NOW 𝑦 =
3
𝑥2+5𝑥+6
• 𝑢 = 3 𝑣 = 𝑥2
+ 5𝑥 + 6
•
𝑑𝑢
𝑑𝑥
= 0
𝑑𝑣
𝑑𝑥
= 2𝑥 + 5
•
𝒅𝒚
𝒅𝒙
=
𝒅
𝒖
𝒗
𝒅𝒙
=
v 𝒅𝒖
𝒅𝒙
− u 𝒅𝒗
𝒅𝒙
𝒗𝟐
• =
(𝑥2+5𝑥+6)(0)−𝟑(𝟐𝒙+𝟓)
(𝑥2+5𝑥+6)𝟐
• =
−𝟑(𝟐𝒙+𝟓)
(𝑥2+5𝑥+6)𝟐
• OR
• =
−𝟔𝒙−𝟏𝟓
(𝑥2+5𝑥+6)𝟐

15. (Find derivative of 𝑥 +
2𝑥+3
𝑥+2
𝑥 +
4𝑥+10
𝑥+3
• 𝑦 = 𝑥 +
2𝑥+3
𝑥+2
𝑥 +
4𝑥+10
𝑥+3
• 𝑦 =
𝑥
1
+
2𝑥+3
𝑥+2
𝑥
1
+
4𝑥+10
𝑥+3
• 𝑦 =
𝑥 𝑥+2 + 2𝑥+3
𝑥+2
𝑥 𝑥+3 + 4𝑥+10
𝑥+3
• 𝑦 =
𝑥2+2𝑥+2𝑥+3
𝑥+2
𝑥23𝑥+4𝑥+10
𝑥+3
• 𝑦 =
𝑥2+4𝑥+3
𝑥+2
𝑥2+7𝑥+10
𝑥+3
• 𝑦 =
𝑥2+3𝑥+𝑥+3
𝑥+2
𝑥2+5𝑥+2𝑥+10
𝑥+3
• 𝑦 =
𝑥 𝑥+3 +1(𝑥+3)
𝑥+2
𝑥 𝑥+5 +2(𝑥+5)
(𝑥+3)
• 𝑦 =
𝑥+1 𝑥+3
𝑥+2
𝑥+5 𝑥+2
𝑥+3
= 𝑥 + 1 𝑥 + 5
• 𝑦 = 𝑥 + 1 𝑥 + 5
• 𝑦 = 𝑥 𝑥 + 5 + 1 𝑥 + 5
• 𝑦 = 𝑥2 + 5𝑥 + 𝑥 + 5
• 𝑦 = 𝑥2
+ 6𝑥 + 5
• NOW
•
𝑑𝑦
𝑑𝑥
=
𝑑 (𝑥2+6𝑥+5)
𝑑𝑥
•
𝑑𝑦
𝑑𝑥
= 2𝑥 + 6

16. Find
𝑑𝑦
𝑑𝑥
if 𝑥 + 1 𝑦 + 1 = 18
• 𝑥 + 1 𝑦 + 1 = 18
• 𝑦 + 1 =
18
𝑥+1
• 𝑦 =
18
𝑥+1
− 1
• 𝑦 =
18−1 𝑥+1
𝑥+1
• 𝑦 =
18−𝑥−1
𝑥+1
• 𝑦=
17−𝑥
𝑥+1
• Now 𝑦 =
17−𝑥
𝑥+1
• LET 𝑢 = 17 − 𝑥 𝑣 = 𝑥 + 1
•
𝑑𝑢
𝑑𝑥
= −1
𝑑𝑣
𝑑𝑥
= 1
•
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
•
𝑑𝑦
𝑑𝑥
=
𝑥+1 −1 −(17−𝑥)(1)
𝑥+1 2
•
𝑑𝑦
𝑑𝑥
=
−𝑥−1−17+𝑥
𝑥+1 2
• =
−18
𝑥+1 2

17. F𝑖𝑛𝑑
𝑑𝑦
𝑑𝑥
𝑖𝑓 𝑥𝑦 + 𝑥 + 𝑦 − 2 = 0
• 𝑥𝑦 + 𝑥 + 𝑦 − 2 = 0
• 𝑥𝑦 + 𝑦 = 2 − 𝑥
• 𝑦 𝑥 + 1 = 2 − 𝑥
• 𝑦 =
2−𝑥
𝑥+1
• N𝑜𝑤 𝑦 =
2−𝑥
𝑥+1
• LET 𝑢 = 2 − 𝑥 𝑣 = 𝑥 + 1
•
𝑑𝑢
𝑑𝑥
= −1
𝑑𝑣
𝑑𝑥
= 1
•
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
•
𝑑𝑦
𝑑𝑥
=
𝑥+1 −1 −(2−𝑥)(1)
𝑥+1 2
• =
−𝑥−1−2+𝑥
𝑥+1 2
• =
−3
𝑥+1 2

18. 𝐴𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠
• If y = 𝑥𝑛 then
𝑑𝑦
𝑑𝑥
=n𝑥𝑛−1
• If y = x then
𝑑𝑦
𝑑𝑥
= 1
• If y=c then
𝑑𝑦
𝑑𝑥
=0
• If y=𝑒𝑥 then
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
• If y= 𝑎𝑥
then
𝑑𝑦
𝑑𝑥
= 𝑎𝑥
loge
a
• 𝐼𝑓𝑦 =loge x then
𝑑𝑦
𝑑𝑥
=
1
𝑥
• 𝐹𝑖𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑥3
𝑒𝑥
𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥
𝐻𝑒𝑟𝑒 𝑦 = 𝑥3
𝑒𝑥
𝑢 = 𝑥2 𝑣 = 𝑒𝑥

𝑑𝑢
𝑑𝑥
= 2𝑥
𝑑𝑣
𝑑𝑥
= 𝑒𝑥

𝑑𝑦
𝑑𝑥
= 𝑣
𝑑𝑢
𝑑𝑥
+ 𝑢
𝑑𝑣
𝑑𝑥
= (𝑒𝑥)(2𝑥)+(𝑥2)(𝑒𝑥)
= 𝑥𝑒𝑥(2 + 𝑥)

19. 𝐴𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠
• If y = 𝑥𝑛 then
𝑑𝑦
𝑑𝑥
=n𝑥𝑛−1
• If y = x then
𝑑𝑦
𝑑𝑥
= 1
• If y=c then
𝑑𝑦
𝑑𝑥
=0
• If y=𝑒𝑥 then
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
• If y= 𝑎𝑥
then
𝑑𝑦
𝑑𝑥
= 𝑎𝑥
loge a
• 𝐼𝑓𝑦 =loge x then
𝑑𝑦
𝑑𝑥
=
1
𝑥
• 𝑓𝑖𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓
𝑥4
log 𝑥
• 𝐻𝑒𝑟𝑒 𝑦 =
𝑥4
log 𝑥
• 𝑢 = 𝑥4
𝑣 = 𝑙𝑜𝑔𝑥
•
𝑑𝑢
𝑑𝑥
= 4𝑥3 𝑑𝑣
𝑑𝑥
=
1
𝑥
•
𝑑𝑦
𝑑𝑥
=
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
=
𝑙𝑜𝑔𝑥 4𝑥3 − 𝑥4 1
𝑥
𝑙𝑜𝑔𝑥 2 =
𝑙𝑜𝑔𝑥 4𝑥3 − 𝑥3
𝑙𝑜𝑔𝑥 2
=
𝑥3 4𝑙𝑜𝑔𝑥−1
𝑙𝑜𝑔𝑥 2

20. 𝐴𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠
• If y = 𝑥𝑛 then
𝑑𝑦
𝑑𝑥
=n𝑥𝑛−1
• If y = x then
𝑑𝑦
𝑑𝑥
= 1
• If y=c then
𝑑𝑦
𝑑𝑥
=0
• If y=𝑒𝑥 then
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
• If y= 𝑎𝑥
then
𝑑𝑦
𝑑𝑥
= 𝑎𝑥
loge a
• 𝐼𝑓𝑦 =loge x then
𝑑𝑦
𝑑𝑥
=
1
𝑥
• 𝐹𝑖𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑥3/2 4𝑥
• 𝑢 = 𝑥3/2
𝑣 = 4𝑥
•
𝑑𝑢
𝑑𝑥
=
3
2
𝑥
3
2
−1
=
3
2
𝑥
1
2
𝑑𝑣
𝑑𝑥
= 4𝑥
log 4
•
𝑑
𝑢
𝑣
𝑑𝑥
=
v 𝑑𝑢
𝑑𝑥
− u 𝑑𝑣
𝑑𝑥
𝑣2
=
𝑙𝑜𝑔𝑥 4𝑥3 − 𝑥4 1
𝑥
𝑙𝑜𝑔𝑥 2 =
𝑙𝑜𝑔𝑥 4𝑥3 − 𝑥3
𝑙𝑜𝑔𝑥 2
=
𝑥3 4𝑙𝑜𝑔𝑥−1
𝑙𝑜𝑔𝑥 2

21. 𝐴𝑝𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠
• If y = 𝑥𝑛 then
𝑑𝑦
𝑑𝑥
=n𝑥𝑛−1
• If y = x then
𝑑𝑦
𝑑𝑥
= 1
• If y=c then
𝑑𝑦
𝑑𝑥
=0
• If y=𝑒𝑥 then
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
• If y= 𝑎𝑥
then
𝑑𝑦
𝑑𝑥
= 𝑎𝑥
loge
a
• 𝐼𝑓𝑦 =loge x then
𝑑𝑦
𝑑𝑥
=
1
𝑥