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STUDENT ID NO
SEAT NO
...
00 p
...
– 2
...
m
...
Show all relevant steps to obtain maximum marks
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912 x 3
...
8190 x 2
...
999
dx

1

sin x 1 

(4 marks)

6

dy
6 cosx 1 

Ans:
dx x 2 sin 7 x 1 
(c) y  ln ln 3 x 
Ans:

(4 marks)

dy
1

dx x ln 3 x 

(d) y  4 ex  e 4 x
Ans:

dy
 4 ex e ln 4  4e 4 x
dx

 

(e) y  cot 1 2 x 3 

Ans:

(4 marks)

1
cot 2 x 3

 

(6 marks)

dy
6x 2
 6 x 2 sec 2 x  6
dx
4x  1

Continued……
...

y
dy
xy ln 10   y
dx x  e xy ln 10 

(5 marks)

(b) Assuming that the following equations define x and y implicitly as differentiable functions x  f t  and

y  g t  , find

dy
at t  
...
Verify it using Quotient Rule
...

NAS

2/3

PEM0036

CALCULUS

9 AUGUST 2017

APPENDIX
BASIC DIFFERENTIATION AND INTEGRATION FORMULAS
𝑑
[sin 𝑥] = cos 𝑥
𝑑𝑥

𝑑
[cos 𝑥] = − sin 𝑥
𝑑𝑥

𝑑
[tan 𝑥] = sec 2 𝑥
𝑑𝑥

𝑑
[sec 𝑥] = sec 𝑥 tan 𝑥
𝑑𝑥

𝑑
[cot 𝑥] = − csc 2 𝑥
𝑑𝑥

𝑑
[csc 𝑥] = −csc 𝑥 cot 𝑥
𝑑𝑥

𝑑 𝑥
[𝑒 ] = 𝑒 𝑥
𝑑𝑥

𝑑
1
[ln 𝑥] = ;
𝑑𝑥
𝑥

𝑥>1

𝑑
1
[𝑠𝑖𝑛−1 𝑥] =
𝑑𝑥
√1 − 𝑥 2

𝑑
1
[𝑐𝑜𝑠 −1 𝑥] = −
𝑓𝑜𝑟 − 1 < 𝑥 < 1
𝑑𝑥
√1 − 𝑥 2

𝑑
1
[𝑡𝑎𝑛−1 𝑥] =
𝑑𝑥
1 + 𝑥2

𝑑
1
[𝑐𝑜𝑡 −1 𝑥] = −
𝑓𝑜𝑟 − ∞ < 𝑥 < ∞
𝑑𝑥
1 + 𝑥2

𝑑
1
[𝑠𝑒𝑐 −1 𝑥] =
𝑑𝑥
|𝑥|√𝑥 2 − 1

𝑑
1
[𝑐𝑠𝑐 −1 𝑥] = −
𝑓𝑜𝑟|𝑥| > 1
𝑑𝑥
|𝑥|√𝑥 2 − 1

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