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
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
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