Maths Study Material

Running head: MATHS
MATHS
Name of the Student
Name of the University
Author Note

Q1 b)
Given, f(x) = (x-2)/sinx
Now, by using the quotient rule for differentiation
( d
dx ) ( f ( x ) ) =
sinx∗( d
dx ) ( x−2 ) – ( x−2 )∗( d
dx ) ( sinx )
sin2 ( x )
= 1∗sinx – ( x−2 ) cos ( x )
sin2 ( x )
Q2
b) y=cos
(−1 ) x = arcos(x)
=> cos(y) = x
Differentiating both sides w.r.t x
 −sin ( y ) ( dy
dx ) =1 (applying product rule of differentiation and as y=f(x)) (1)
 dy
dx = −1
sin ( y ) =¿ −1
√ ( 1−x2 ) (as sin^(θ) + cos^(θ) = 1)
Again differentiating (1) w.r.t x
d2 y
d x2 sin ( y ) + (( dy
dx )
2
) cosy = 0
 d2 y
d x2 sin ( y ) = −( ( dy
dx )
2
) cosy = −1
1−x2 ∗x
 d2 y
d x2 √1−x2 = −1
1−x2 ∗x
 d2 y
d x2 =
−x
( 1−x2 )
3
2

c) f ( x ) =sinh ( −1 ) ( 5 x )
sinh(y) = (e^(y) - e(-y))/2 (By definition of hyperbolic function)
Now, y = f(x) = arcsinh(5x)
 sinh(y) = 5x
 (e^(y) - e(-y))/2 = 5x
 (e^(y) - e(-y)) = 10x
 (dy/dx)* (e^(y) + e(-y)) = 10 (derivating both sides w.r.t x)
 (dy/dx)* (e^(y) + e(-y))/2 = 5
 (dy/dx)*cosh(y) = 5
 (dy/dx)* √1+sinh ( y )2 = 5 (as cosh (θ)2 – sinh(θ)2=1)
 (dy/dx) = 5
√1+sinh ( y )2 = 5
√1+25 x2 (as sinh(y) = 5x)
Q3:
a) ∫ 1
2 ( 5 x−3 ) 6 dx,
let, ( 5 x−3 ) =z => 5dx = dz (taking natural derivative on both sides)
 dx = dz/5
Hence, ∫ 1
2 ( 5 x−3 )6 dx = ∫ 1
10 z6 dz = z7
70 + C = ( 5 x−3 ) 7
70 +C
Q4:
a)
i=50 sin(100 πt) t>=0
i = current in mA, t = time in seconds.