Calculus WIs

Answers to selected exercises from 04_HW

(𝑥 𝑝 )′ = 𝑝𝑥 𝑝−1 , 𝑝∈𝑅
1 1 1
(ln 𝑥)′ = , (log 𝑎 𝑥)′ =
𝑥 𝑥 ln 𝑎
(𝑒 𝑥 )′ = 𝑒 𝑥 , (𝑎 𝑥 )′ = 𝑎 𝑥 ⋅ ln 𝑎
1
(sin 𝑥)′ = cos 𝑥 , (cos 𝑥)′ = − sin 𝑥 , (tan 𝑥)′ =
cos 2 𝑥
1 −1
(𝑎𝑟𝑐 sin 𝑥)′ = , (𝑎𝑟𝑐 cos 𝑥)′ = ,
√1 − 𝑥 2 √1 − 𝑥 2
1 −1
(𝑎𝑟𝑐 tan 𝑥)′ = , (𝑎𝑟𝑐 ctan 𝑥)′ =
1 + 𝑥2 1 + 𝑥2

1. (𝑐𝑓 ′ ) = 𝑐𝑓 ′ , (𝑓 ± 𝑔)′ = 𝑓 ′ ± 𝑔′

2. (𝑓 ⋅ 𝑔)′ = 𝑓 ′ 𝑔 + 𝑓𝑔′

𝑓 ′ 𝑓 ′ 𝑔 − 𝑓𝑔′
3. ( ) =
𝑔 𝑔2
′
4. ( 𝑓(𝑦(𝑥)) ) = 𝑓 ′ (𝑦) ⋅ 𝑦 ′ (𝑥)
′ ′
5. 𝑓 ′′ (𝑥) = (𝑓 ′ (𝑥)) , 𝑓 𝑛 (𝑥) = ( 𝑓 𝑛−1 (𝑥))

𝜋
sin(𝑥)−sin( )
7
0. 𝑎) lim𝜋 𝜋 = 𝑎. It is easily seen that this is the definition f the derivative sin 𝑥, at 𝑥0 =
𝑥→ 𝑥−
7 7
𝜋 𝜋
7
, so 𝑎 = cos 7 .

𝑓(𝑥)−3
b) Suppose that 𝑥0 = 1 belongs to the domain of 𝑓(𝑥). If lim 𝑥−1
= 3, find 𝑓 ′ (1).
𝑥→1

First we note that 𝑓(1) = 3, 2. Next from the definition of the derivative we obtain, that 𝑓′(1) = 3.

1. Calculate the following simple derivatives

(𝑥 𝑝 )′ = 𝑝𝑥 𝑝−1 , 𝑝∈𝑅

𝑎) (𝑥 4 )′ = 4𝑥 3

𝑏) (7)′ = 0

𝑐) (3𝑠)′ = 3

1
 7
𝑑) (𝑥 −7 )′ = −7𝑥 −8 = −
𝑥8

′ 1 ′ 1 1 1
𝑒) (√𝑡) = (𝑡 2 ) = 𝑡 −2 =
2 2√𝑡

′ 1 ′ 1 −3 1 1
𝑓) ( 4√𝑥 ) = (𝑥 4 ) = 𝑥 4 = 4
4 4 √𝑥 3

3 ′ 4 ′ 4 1 43
𝑔) ( √𝑥 4 ) = (𝑥 3 ) = 𝑥 3 = √𝑥
3 3

1 ′ −1
ℎ) ( ) = (𝑦 −1 )′ = −𝑦 −2 = 2
𝑦 𝑦

𝑖) 𝑓 ′ (𝑦) = 0 , the function f is a function of y, x does not change, so also 1/𝑥 3 does not change
with respect to y.

1 ′ 2 ′ 2 5 2 1
𝑗) ( 3 ) = (𝑥 −3 ) = − 𝑥 −3 = − 3
√𝑥 2 3 3 √𝑥 5

𝑎 ′ 1 ′ 7
𝑘) ( 8 ) = ( 7 ) = ( 𝑎 −7 )′ = −7𝑎−8 = − 8
𝑎 𝑎 𝑎
′
𝑥4 8 ′ 4 ′ 4 1 43
𝑙) ( 3 ) = (𝑥 4−3 ) = (𝑥 3 ) = 𝑥 3 = √𝑥
√𝑥 8 3 3

2. Find the derivatives of the following functions.

e − (ln t )et
1 t 1
b) f (t ) = 
 ln t  (ln t ) e − (ln t ) e
'

=
' t
( )t '
= t = t
− ln t
t 
( )
'

e  et ( )
2
et
2
( ) et

   1
'
  (2 z )− 3  2 + 2 z .
c) f (z ) =  3
1 1 2
'
 = −
 2 z + z  
2
(
3
2
)
2 z + z 2  3 

   1
'
  (2 )− 3  2 + 2 x .
d) f ( ) =  3
1 1 2
'
 = −
2 
 2 + x   ( 3
2 + x 2  3
2
) 

'

g) f ' ( y ) =  5 tan + tan  = 5
y 1 1 1
 +0=
 5 8  y 5  y
cos2   cos2  
5 5

2
   1 
 1 +  
'
    2 x 
h) f (x ) =  x + x + x  =
1
'
 1 + 
  2 x+ x+ x  2 x+ x 
 
 

 1 
 − 
'
− −  − x − (1 − ln x ) 
      x
i) y' = f ' (x ) =  sin 2 
1 ln x 1 ln x 1 ln x 
  = 2 sin   cos 
  x   x   x  x2 
 
 
2 +𝑙𝑛 𝑥+1 1 2 1
′ 2 +𝑙𝑛 𝑥+1 ′ 𝑒𝑥 (2𝑥 + 𝑥 ) √𝑥 − 𝑒 𝑥 +𝑙𝑛 𝑥+1
1 𝑥 2 +𝑙𝑛 𝑥+1
𝑒𝑥 2√𝑥
𝑘) 𝑓 ′ (𝑥) = ( 𝑒 ) =( ) = =
√𝑥 √𝑥 𝑥

2 +1 1 1
𝑒𝑥 𝑥 [(2𝑥 + 𝑥 ) √𝑥 − ]
2√𝑥 2 1 1
= = 𝑒 𝑥 +1 [(2𝑥 + ) √𝑥 − ]
𝑥 𝑥 2√𝑥
′
𝑙) 𝑦′ = 𝑓 ′ (𝑥) = (𝑒 𝑠𝑖𝑛 𝑥 ) = 𝑒 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥

3. Find the derivatives of [hint: use the equality x = e ln x ]

𝑎) 𝑦 = 𝑥 𝑥 ;

𝑥 ′ ′ 1
(𝑥 𝑥 )′ = (𝑒 ln(𝑥 ) ) = (𝑒 𝑥⋅ln(𝑥) ) = 𝑒 𝑥⋅ln(𝑥) (𝑥 ⋅ ln(𝑥))′ = 𝑒 𝑥⋅ln(𝑥) (ln 𝑥 + 𝑥 )
𝑥
𝑥⋅ln(𝑥) (ln 𝑥
=𝑒 𝑥 + 1) = 𝑥 (ln 𝑥 + 1)

b) y = sin x x

𝑥 ′ ′ 𝑐𝑜𝑠 𝑥
𝑓 ′ (𝑥) = ((𝑠𝑖𝑛 𝑥)𝑥 )′ = (𝑒 𝑙𝑛((𝑠𝑖𝑛 𝑥) ) ) = (𝑒 𝑥 𝑙𝑛(𝑠𝑖𝑛 𝑥) ) = 𝑒 𝑥 𝑙𝑛(𝑠𝑖𝑛 𝑥) (𝑙𝑛(𝑠𝑖𝑛 𝑥) + 𝑥 ⋅ )=
𝑠𝑖𝑛 𝑥

= 𝑒 𝑥 𝑙𝑛(𝑠𝑖𝑛 𝑥) (𝑙𝑛(𝑠𝑖𝑛 𝑥) + 𝑥 ⋅ 𝑐𝑜𝑡 𝑥) = (𝑠𝑖𝑛 𝑥)𝑥 (𝑙𝑛(𝑠𝑖𝑛 𝑥) + 𝑥 ⋅ 𝑐𝑡𝑔𝑥)

c) y = x
sin x

( '
) ( sin x
) (
' 
) 1
f ' (x ) = x sin x = e ln (x ) = e sin x ln x = e sin x ln x  cos x ln x + sin x   =
'

 x

 1
= x sin x  cos x ln x + sin x  
 x

e)
f ( x) = (cos3 x) cot x e 2 x

( ) ( ) ( )
f ' (x ) = (cos3 x) cot x e 2 x = (cos3 x) cot x e 2 x + e 2 x (cos3 x) cot x =
' ' '

3
(e ( ln (cos3 x )cot x
) (
) ' e 2 x + 2e 2 x (cos3 x) cot x = ecot x ln (cos x ) ' e 2 x + 2e 2 x (cos3 x) cot x =
3
)
( 3  1
 sin x
)
= e cot x ln (cos x )  − 2 ln cos3 x + cot x
1
3
cos x
( )

 3 cos2 x  (− sin x )e 2 x + 2e 2 x (cos3 x) cot x =


 ln cos3 x
= (cos3 x) cot x e 2 x  − − 3 + 2

 = −
(
(cos3
x ) e
)
cot x 2 x  ln cos x

3

+ 1
( )
2   2
 sin x   sin x 
1
f) 𝑓(𝑦) = (cos 𝑦)𝑦

1 ′ 1 ′ 1 1 ′
′(𝑦) ln (cos 𝑦)⋅ ln cos 𝑦⋅
𝑓 = 𝑦
((cos 𝑦) ) = (𝑒 𝑦) =𝑒 𝑦 (ln (cos 𝑦) ⋅ )
𝑦
1 −sin 𝑦 1 −1
ln cos 𝑦⋅
=𝑒 𝑦 ( + ln cos 𝑦 2 ).
cos 𝑦 𝑦 𝑦

d2y
4. Find 𝑦" or for
dx 2
a) y = x 1 + x 2

(
f ' (x ) = x 1 + x 2 = 1 + x 2 + x )
'

2 1+ x2
1
 2x = 1 + x 2 +
x2
1+ x2
=
1 + 2x 2
1+ x2

 1 + 2x 
' 4x 1 + x 2 − 1 + 2x 2  ( ) 1
 2x (
4x 1 + x2 − 1 + 2x2  ) x
2 1 + x2 1 + x2 =
2
f '' (x ) =   =
 =
 1+ x
2
 1 + x2 1 + x2
𝑥
4𝑥√1 + 𝑥 2 − (1 + 2𝑥 2 ) ⋅
√1 + 𝑥 2 4𝑥(1 + 𝑥 2 ) − (1 + 2𝑥 2 ) ⋅ 𝑥 4𝑥 + 4𝑥 3 − 𝑥 − 2𝑥 3
= = =
1 + 𝑥2 (1 + 𝑥 2 )√1 + 𝑥 2 (1 + 𝑥 2 )√1 + 𝑥 2
3
2𝑥 + 3𝑥
=
(1 + 𝑥 2 )√1 + 𝑥 2

2)
b) 𝑦 = 𝑒 (−𝑥
2 ′ 2 2)
𝑓 ′ (𝑥) = (𝑒 (−𝑥 ) ) = 𝑒 (−𝑥 ) ⋅ (−2𝑥) = −2𝑥 ⋅ 𝑒 (−𝑥

2 ′ 2 2 2 2)
𝑓 ′′ (𝑥) = (−2𝑥 ⋅ 𝑒 (−𝑥 ) ) = −2𝑒 (−𝑥 ) − 2𝑥𝑒 (−𝑥 ) ⋅ (−2𝑥) = −2𝑒 (−𝑥 ) + 4𝑥 2 𝑒 (−𝑥
2
= 𝑒 (−𝑥 ) (4𝑥 2 − 2)

d2y
5. Let f (x) be two times differentiable. Find y' ' or for
dx2
a) y = f ( x )
2

4
 ′
𝑎) 𝑦 ′ = (𝑓(𝑥 2 )) = 𝑓 ′ (𝑥 2 )2𝑥, 𝑦 ′′ = (𝑓 ′ (𝑥 2 )2𝑥)′ = 𝑓 ′′ (𝑥 2 ) ⋅ 2𝑥 ⋅ 2𝑥 + 2𝑓 ′ (𝑥 2 ).

( )'
y ''' = 2  f ' ( x 2 ) + 4 x 2  f '' ( x 2 ) = 2  f '' ( x 2 )  2 x + 8x  f '' ( x 2 ) + 4 x 2  f ''' ( x 2 )  2 x =

= 12x  f '' ( x 2 ) + 8 x 3  f ''' ( x 2 )

b) y = f (1 / x)

y ' = ( f (1 / x) ) = −
1
 f ' (1 / x)
'
2
x
′′
𝑦 ′′ = (𝑓(𝑥 −1 )) = (−𝑓 ′ (𝑥 −1 ) ⋅ 𝑥 −2 )′ = 𝑓 ′′ (𝑥 −1 ) ⋅ 𝑥 −2 ⋅ 𝑥 −2 + 𝑓 ′ (𝑥 −1 )2𝑥 −3 lub
'
 1  2  1  1 2 1
y '' =  − 2  f ' (1 / x)  = 3  f ' (1 / x) +  − 2    − 2   f '' (1 / x) = 3  f ' (1 / x) + 4  f '' (1 / x)
 x  x  x   x  x x
'
 2 1 
y =  3  f ' (1 / x) + 4  f '' (1 / x)  =
'''

x x 

6 ' 2  1  4 1  1 
− f (1/ x) + 3  f '' (1/ x)   − 2  +  − 5   f '' (1/ x) + 4  f ''' (1/ x)   − 2  =
 x   x   x 
4
x x x

6 ' 6 1
− 4
f (1/ x) − 5  f '' (1/ x) − 6  f ''' (1/ x)
x x x

c) y = f (ln x)

′′ ′ (ln
1 ′ 1 1
(𝑓(ln 𝑥)) = (𝑓 𝑥) ) = 𝑓 ′′ (ln 𝑥) 2 − 𝑓 ′ (ln 𝑥) 2 .
𝑥 𝑥 𝑥

The equation of a line tangent to f (x) at (x0, f (x0)):

𝒚 = 𝒇(𝒙𝟎 ) + 𝒇′(𝒙𝟎 )(𝒙 − 𝒙𝟎 )

6. Find the equation of the tangent line to the given function 𝑓 (i.e. the linear function
which is the approximation of 𝑓 ) at the indicated point

𝑏) 𝑓(𝑥) = 2𝑥 2 + 3𝑥 − 7 at (−1, −8) Df = R

𝑓′(𝑥) = (2𝑥 2 + 3𝑥 − 7)′ = 4𝑥 + 3

𝑓′(−1) = −4 + 3 = −1

𝑦 = −8 − 1(𝑥 + 1)

𝑦 = −8 − 𝑥 − 1 𝑡ℎ𝑢𝑠 𝑦 = −𝑥 − 9

Ans: The tangent line to the graph of 𝑓(𝑥) = 2𝑥 2 + 3𝑥 − 7 at (−1, −8) is 𝑦 = −𝑥 − 9.

5
7. a) Determine the equation of the tangent line to the curve y = x 2 + 2 x at point (1, 3).
𝑥0 = 1

𝑓(𝑥0 ) = 3

𝑓(𝑥) = 𝑥 2 + 2𝑥

𝑓 ′ (𝑥) = 2𝑥 + 2

𝑓 ′ (𝑥0 ) = 4

𝑦 = 3 + 4(𝑥 − 1)

𝑦 = 3 + 4𝑥 − 4

𝑦 = 4𝑥 − 1

Ans.: The tangent to 𝑓(𝑥) = 𝑥 2 + 2𝑥 at (1,3) is 𝑦 = 4𝑥 − 1

2𝑥
𝑓 ′ (𝑥) = − (1+𝑥 2)2

1
𝑓 ′ (−1) =
2
1 1 1 1 1 1
𝑦= + (𝑥 + 1) ⇒ 𝑦= + 𝑥+ ⇒ 𝑦= 𝑥+1
2 2 2 2 2 2
1 1
Ans.: The tangent to 𝑓(𝑥) = 1+𝑥 2 at (−1, 1/2) is 𝑦 = 2 𝑥 + 1.

WolframApha: tangent to 1/(1+x^2) at x=-1

𝑥
𝑐) 𝑆𝑒𝑟𝑝𝑒𝑛𝑡𝑖𝑛𝑒 𝑓(𝑥) = , 𝑥 = −3, 𝑓(𝑥0 ) = −0.3
𝑥2 +1 0

6
 𝑥 2 + 1 − 𝑥 ⋅ 2𝑥 1 − 𝑥2
𝑓 ′ (𝑥) = =
(𝑥 2 + 1)2 (𝑥 2 + 1)2

8 2
𝑓 ′ (−3) = − =−
100 25
2 2 27
𝑦 = −0.3 − 25 (𝑥 + 3) = − 25 𝑥 − 50 - tangent line.