Mathematics 1A Assignment 4

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Added on  2020/05/11

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This Mathematics 1A assignment consists of ten questions covering various calculus concepts. Students need to apply the product rule, quotient rule, and chain rule to find derivatives. They will also utilize the mean value theorem and L'Hopital's rule to solve limits. The assignment includes problems on optimization, proving theorems, and understanding the behavior of functions.

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300672 MATHEMATICS 1A
ASSIGNMENT – 4
STUDENT ID
[Pick the date]

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Question 1
(a) y=cosh2( x ¿¿ 5) sinhx ¿
Let
f ( x )=cosh2( x ¿¿ 5)¿
g ( x )=sinhx
Apply product rule
( f . g ) ' =f ' . g+ f . g'
dy
dx = d
dx {cosh2 (x ¿¿ 5)}sinh ( x)+ d
dx {sinh (x )}cosh2( x¿¿ 5) (a)¿ ¿
= I + II
Solve I
I = d
dx {cosh2 (x ¿¿ 5)}¿
df ( u )
dx = df
du . du
dx
Let
cosh ( x ¿¿ 5)=u ¿
¿ d
du ( u2 ) . d
dx {cosh2 (x ¿¿ 5)}¿
¿ 2 u .sinh ( x¿ ¿5).5 x4 ¿
Substitute u=cosh ( x¿ ¿5)¿
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¿ 2 cosh (x¿ ¿5). sinh(x ¿¿ 5). 5 x4 ¿¿
Simplify
I =10 x4 cosh ( x¿ ¿5) sinh(x ¿¿ 5)¿ ¿
Now,
II= d
dx {sinh (x )}
II=cosh ( x )
Now, substitute I and II in equation (a)
dy
dx =10 x4 cosh ( x ¿¿ 5)sinh( x ¿¿ 5)sinh ( x ) +cosh ( x ) cosh2 ( x¿ ¿5)¿ ¿ ¿
(b) y= sinh( x)
cos ( x )
Let
f =sinh ( x )
g=cos(x )
Applying the Quotient Rule
( f
g )'
= f ' . gg' . f
g2
¿ d
dx ¿ ¿
¿ ¿ ¿
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¿ cosh ( x ) .cos ( x ) +sin (x) sinh (x )
cos2 (x)
Question 2
d
dx ( tanh1 x ) = 1
1x2
Let
y=tanh1 x
Or
x=tanh y
Differentiate both side w.r.t. x
d
dx tanh y = d
dx ( x )
sech2 y dy
dx =1
dy
dx = 1
sech2 y ( 1 )
sech2 y=1tanh2 y
dy
dx =
( 1
1tanh2 y )
x=tanh y
3

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dy
dx =
( 1
1x2 )
Put
y=tanh1 x
d
dx (tanh ¿¿1 x)= ( 1
1x2 ) ¿ Proved
Question 3
Absolute maximum and minimum value of f ( x)=?
f ( x )=4 x
2
7 + x
9
7 , closed interval [ 1 , 1 ]
Now,
f ( x )=4 x
2
7 + x
9
7
Take first derivative of both the sides
f ' ( x ) =4 × ( 2
7 ) x
2
7 1
+( 9
7 ) x
9
7 1
f ' ( x ) = 8
7 x
5
7 + 9
7 x
2
7
Take f ' ( x )=0
f ' ( x )= 8
7 x
5
7 + 9
7 x
2
7 =0
8
7 x
5
7 + 9
7 x
2
7 =0
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8
7 x
5
7 =9
7 x
2
7
8 x
5
7 =9 x
2
7
8
x
5
7
=9 x
2
7
8
9 =x
2
7 . x
5
7
8
9 =x
7
7
x= (8
9 )
Further,
f ( x )=4 x
2
7 + x
9
7
At x= (8
9 )
f (8
9 )= {4(8
9 )2
7
}+ (8
9 )9
7
f (8
9 )=3.008
At x=0
f (0)=0 (critical point since f ' (0)is undefined)
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At x = 1
f ( 1 ) =4 +1=5
At x = -1
f (1 )=3
Therefore, function f(x) has absolute maximum value is 5 at x=1and absolute minimum value is
0 at x = 0.
Further, f has local maximum value is 3.008 at x = -8/9.
Question 4
f ( x )=ln( x2+ 4 x +8)
(a) F(x) is increasing or decreasing
First derivation of function f(x)
dy
dx = d
dx (ln ( x2 +4 x +8 ) )
¿ ( 1
x2 +4 x +8 ) d
dx ( x2 +4 x+8 )
¿ 1
x2 + 4 x+8 ( 2 x+4 )
¿ 2 x+ 4
x2 + 4 x+ 8
f ' ( x )= dy
dx = 2 x+ 4
x2 + 4 x+ 8
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Put f’(x) = 0
2 x+ 4
x2 + 4 x+8 =0
2 x+ 4=0
x=2
Sign of f’(x) would be given below:
f '> 0: f isincreasing on (2 , )
f '< 0: f is decreasingon (,2 )
(b) Values x for which function f(x) would concave up and concave down
Second derivative of the function is shown below:
f ' ( x )= dy
dx = 2 x+ 4
x2 + 4 x+ 8
f '' ( x )= d2 y
d x2 = d
dx ( 2 x +4
x2+ 4 x +8 )
¿( x ¿¿ 2+ 4 x+ 8). d
dx ( 2 x + 4 ) ( 2 x +4 ) . d
dx
( x ¿¿ 2+4 x +8)
(x ¿¿ 2+4 x+ 8)2 ¿ ¿ ¿
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¿ ( x ¿¿ 2+4 x +8) ( 2 ) ( 2 x+4 ) ( 2 x+ 4 )
( x ¿¿ 2+ 4 x +8)2 ¿ ¿
¿ (2 x¿ ¿2+8 x +16) ( 4 x2 +16 x+16 )
(x ¿¿ 2+4 x +8)2 ¿ ¿
¿ 2 x2 +8 x+164 x216 x16
(x ¿¿ 2+4 x+ 8)2 ¿
¿ 2 x28 x
(x ¿¿ 2+4 x +8)2 ¿
f ' ' ( x ) = 2 x (x+ 4)
(x ¿¿ 2+4 x +8)2 ¿
Now, put f ' ' ( x )=0
2 x ( x+4 )
( x ¿¿ 2+4 x +8)2=0¿
2 x ( x +4 )=0
x=0 ,4
Sign of f’(x) would be given below:
f ' ' >0: concave up on ( ,4 ) (0 , )
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f '< 0: concave down on (4 , 0 )
Question 5
f ' ( x )= dy
dx =(x 5)5 (x 6)6 ( x7)7 ( x8)8 ( x9)9
F(x) has local maximum and minima
f ' ( x )= dy
dx = ( x5 )5 ( x6 )6 ( x7 )7 ( x8 )8 ( x9 )9=0
( x5 )5 ( x6 )6 ( x7 )7 ( x8 )8 ( x9 )9=0
x=5 , 6 , 7 , 8 , 9
The values of x would be termed as critical number of the given functions. These critical
numbers would be exists if the respective first derivation would be undefined at the values of x.
After determining the peaks and valley it would be fair to conclude that (4, -6) would be the local
minima. However, these two local extreme would be plugged in the original function. Hence, the
local maxima have been found as (-4, 7).
Question 6
Sketch a curve as y = f(x) for the following conditions is represented below:
9

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Question 7
Let
P ( a , b )Q ( c , d ) arethe two points on the given cubic
y=x3 + A
Where, 0<a<c
By using the mean value theorem that a point R (r, s) on PQ is such that tangent at R is || to chord
PQ.
y=x3 + A
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dy
dx =3 x2
Let r ( a , c ) , a<r < c
Mean value theorem
dy
dx |x=r
= f ( c ) f ( a )
ca
f ( c ) =c3+ A
f ( a ) =a3+ A
dy
dx |x=r
=3 r2
Hence,
3 r2= ( c3 + A ) (a3 + A )
ca
3 r2= c3 + Aa3 A
ca
3 r2= c3a3
ca
3 r2= ( ca ) (c ¿¿ 2+ac +a2)
ca ¿
3 r2=c2 +ac + a2
r2= c2 +ac +a2
3
r = ( c2+ ac+ a2
3 )
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Proved
It can also be concluded that point R(r , s) is also exists on arc PR such that tangent at R is || to
the chord PQ.
Question 8
(a) lim
x 0
x
sinhx
lim
x 0
( x
sinh ( x ) ¿)¿
Apply L’Hopital’s Rule
¿ lim
x 0
( 1
cosh ( x ) ¿)¿
Put x=0
¿ 1
cosh ( 0 )
¿ 1
(b) lim
x
ln ( lnx )
lnx
lim
x
ln ( lnx )
lnx
Apply L’Hopital’s Rule
12

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¿ lim
x
1
x ln (x )
1
x
¿ lim
x
1
ln ( x )
lim
x a
a [ f (x )
g ( x ) ] =
limf (x)
x a
lim
x a
g( x ) , lim
x a
g ( x ) 0
** With the exception of the indeterminate form
¿
lim
x
(1)
lim
x
¿ ¿
Now,
lim
x
( 1 ) =1
lim
x
¿
¿ 1

Applying the infinity properties a
=0
¿ 0
(c) lim
x 0+¿ ln x sinx ¿
¿
lim
x 0+¿ ln(x)sin ( x)¿
¿
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Indeterminate form, rewrite to accommodate for L’Hopital’s form
a . b= a
1
b
, b 0
ln ( x ) sin ( x ) = ln ( x )
1
sin ( x )
¿ lim
x 0+¿
( ln ( x )
1
sin ( x ) ) ¿
¿
Apply L’Hopital’s Rule
¿ lim
x 0+¿ ( 1
x
cot ( x ) cosine ( x ) ) ¿
¿
¿ lim
x 0+¿ ( sin ( x)
x cot (x) ) ¿
¿
lim
x a
[ c . f ( x ) ] =c . lim
x a
f ( x )
lim
x a
a [ f (x )
g ( x ) ] =
limf (x)
x a
lim
x a
g( x ) , lim
x a
g ( x ) 0
** With the exception of the indeterminate form
¿ lim
x 0+¿sin ( x )
¿ lim
x 0+¿(x cot ( x ) )
¿ ¿ ¿
lim
x 0+¿ sin ( x ) =0 ( when puting x=0 ) .(1)¿
¿
lim
x 0+¿ ( xcot (x ) ) ¿
¿
Apply L’Hopital’s Rule
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cot ( x ) x= x
( 1
cot ( x ) )
lim
x 0+¿ x
( 1
cot ( x ) ) ¿
¿
Apply L’Hopital’s Rule
lim
x 0+¿ ( cos2 ( x ) ) ¿
¿
¿ cos2 ( 0 ) =1 ( when puting x=0 ) ..(2)
Now,
¿ lim
x 0+¿sin ( x )
¿ lim
x 0+¿(x cot ( x ) )
¿ ¿ ¿
From equation (1) and (2)
¿ 0
1
¿ 0
Question 9
lim
x 0+¿ xsin x
(1cosx )
3
2
¿
¿
¿ cos x=12 sin2 x /2
¿ lim
x 0+¿ xsin x
(1(12 sin2 x/2 ))
3
2
¿
¿
15

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¿ lim
x 0+¿ xsin x
(11+2 sin2 x/ 2)¿
3
2
¿
¿ lim
x 0+¿ xsin x
(2 sin2 x/2 )¿
3
2
¿
¿ lim
x 0+¿ xsinx
2 2 sin3
x/ 2 ¿
¿
¿ lim
x 0+¿ 1cosx
2 2 . 3
2 sin2
( x
2 )cos ( x
2 ) ¿
¿
¿ lim
x 0+¿ 1cosx
3 2 sin2
( x
2 ) cos ( x
2 ) ¿
¿
¿ lim
x 0+¿ (sinx )
3 2 {sinx cos ( x
2 )sin3
( x
2 )
2 }¿
¿
¿
¿
¿ lim
x 0+¿ sinx
3 2 {sinx cos ( x
2 ) sin3
( x
2 )
2 2 }¿
¿
¿
¿
¿ lim
x 0+¿ 1
3 {cos ( x
2 )sin3
( x
2 )
2 }¿
¿
¿
¿
¿ lim
x 0+¿ 2
3cos ( x
2 )sin3
( x
2 )
¿
¿
¿
¿
¿ lim
x 0+¿ 2
3 { 1
cos ( x
2 )sin3
( x
2 ) }¿
¿
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¿ 2
3 ( 1
10 )
¿ 2
3
Question 10
The function f(x) is such that
f ( 3 )=9
f ( 5 )=15
Further,
f ' ( x ) 3
Needs to prove that f(x) = 3x for all 3 x 5
Mean value theorem
f ' ( x ) = f ( b ) f ( a )
ba
Let c [ 3,5 ]
f ' ( x ) = f ( b ) f ( a )
ba
f ' ( x ) = f ( 5 ) f ( 3 )
(53)
f ' ( x )= 159
2
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f ' ( x )= 6
2
f ' ( x )=3
Take integral of both the side w.r.t. x
f ' ( x ) =3
d
dx f ( x ) = 3
d f ( x )=3 dx
f ( x )=3 x { for all 3 x 5 }
Proved
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