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MATH 320
Calculus II
Differentiation review exercises solutions
Differentiate
1. f ( x)  13x10  17 x5  19 x  23


   
Note that this verifies  130 x  85 x
f ( x)  13 10 x9  17 5x 4  19 1x0  0  130 x9  85x 4  19
9
2. g (t )  6 t 
4

 19 dx  13x10  17 x 5  19 x  C .
1
6 t
First, use algebra (properties of exponents) to rewrite g (t )  6t 1 / 2  16 t 1 / 2 .


g t   6  12 t 1 / 2  16  12 t 3 / 2  3t 1 / 2  121 t 3 / 2
3
1
3
1
Optional, rewrite as g t   1 / 2 
or as g t  
.

3/ 2
t
12t
t 12t t
3. y 
t 2 t  5t  25
t
Solution 1 (not shown): use quotient rule, noting t 2 t  t 5 / 2 .
Solution 2: first use algebra to rewrite y  t 3 / 2  5  25t 1 , then differentiate to obtain
dy 3 1 / 2
dy 3
25
 2 t  0  25 t  2 ; optionally rewrite as
2 t 2 .
dx
dx
t
4. k ( x)  sec x tan x

Product rule uv   uv  uv :
k x   sec x sec 2 x  sec x tan x  tan x  sec3 x  sec x tan 2 x


Optional: use algebra and trigonometric identity tan 2 x  1  sec 2 x to rewrite this as
k x   sec x sec 2 x  tan 2 x  sec x 1  2 tan 2 x or k x   sec x 2 sec 2 x  1 .
5. h( x)  e x sin( e x )





Product rule, then chain rule – if u is differentiable function of x, then
   e   e
hx   e x cos e x
d
dx
x
x
   
2
 
 

d
dx
sin u   cos u dudx :
 
 
sin e x  e x cos e x  e x sin e x  e 2 x cos e x  e x sin e x
6. j (t )  sin 2 t  cos2 t
Solution 1 (from class): first use the trigonometric identity sin 2 t  cos 2 t  1 , thus j (t )  1
and jt   0 .
Solution 2 (derivative of sum, then chain rule on each term; note sin 2 t  sin t  ):
j(t )  2 sin t dtd sin t   2 cos t dtd cos t   2 sin t cos t  2 cos t ( sin t )  2(sin t cos t  cos t sin t )  0
2
7. m( s)  49s  19
39
Chain rule (power rule form – if u is differentiable function of x, then


d
dx
38
u   nu
n
38
38
ms   4 399s  19 dsd 9s  19  4  39  99s  19  14049s  19
n 1 du
dx
):
8. n( x)  3 3 x  cot x
Use algebra to rewrite as n x   3 x  cot x 
1/ 2
nx  
1
3
3x  cot x  2 / 3 3   csc 2 x  
, then use power rule form of chain rule:
3  csc 2 x
2/3
33x  cot x 
 t2 1 

9. s  ln  3
t

2
t


Solution 1 (first use properties of logarithms to rewrite, then use chain rule,
d
ln u   1 du ):
dx
u dx
ds
1 d 2
1 d 3
2t
3t 2  2
 2
t

1

t

2
t


s  ln t 2  1  ln t 3  2t , so
dt t  1 dt
t 3  2t dt
t 2  1 t 3  2t
Optional: use algebra to rewrite as
ds 2t t 3  2t  3t 2  2 t 2  1 2t 4  4t 2  3t 4  5t 2  2
 t4  t2  2



dt
t 2  1 t 3  2t
t 2  1 t 3  2t
t 2  1 t 3  2t
du
Solution 2 (see #10) – use chain rule, and quotient rule to find
.
dt



10. z 

 
 






x2  1
x3  2 x
2





 






 u  vu  uv
Quotient rule   
:
v2
v
dz
x3  2 x 2 x   x 2  1 3x 2  2
and optionally, simplify to

2
dx
x3  2 x







dz 2 x3  4 x 2  (3x 4  5 x 2  2) ( x 4  x 2  2)
.


2
2
dx
x3  2 x
x3  2 x




t2 1
and change variable x to t in #10:
t 3  2t
ds 1 du
1
 t4  t2  2
 t4  t2  2
, the same answer we obtained in

 2

2
t 1
dt u dt
t 2  1 t 2  2t
t 3  2t
t 3  2t
solution 1.
Solution 2 for #9, let s = ln u, where u 

 
 




11. y  cos x  x sin x  e
dy
  sin x  x cos x  sin x   0  x cos x
dx
12. g ( x) 
x cos( x 4 )
2x4  
Use quotient rule, then for derivative of numerator, use product then chain rule:
2 x 4   dxd x cos x 4  x cos x 4 dxd 2 x 4  
g x  
2
2x4  

2x

4
2x
4

13. Find
 
 
 
  


    
2x   
sin x   cosx   8x
2x   
  
  x  sin x 4 4 x3  cos x 4  x cos x 4 8 x3
2
4

   4x4
4
4
dy
dx

4
4
 
cos x 4
2
: xy4  x 4 y  csc x
Differentiate implicitly with respect to x (on left, use difference, then product and chain):
d
d
csc x
xy4  x 4 y 
dx
dx
dy
dy
Now, solve for dy
:
x 4 y3 dx  1y 4  x4 dx  4 x3 y   csc x cot x
dx




 

3
4xy
3
 x4

dy
dx
  csc x cot x  y 4  4 x3 y ,
dy 4 x3 y  csc x cot x  y 4
(for (x, y) satisfying 4 xy3  x 4  0 and both csc x

3
4
dx
4 xy  x
and cot x are defined.
so
End notes:
 Similarly as in #1, we have verified the indefinite integral of our answer equals the given
function plus a constant, C, of integration.
 Using integration rules in section 5.1, we could evaluate  130 x9  85 x 4  19 dx. , the










answer in problem 1.
If we first used algebra, or a trigonometric identity, we could integrate the “answers” in
problems 2 and 3.
The intent of problem 6 is to advise you to be on the lookout for using identities.
To integrate functions where the chain rule is used to check the answer, we’ll later use
the “method of substitution”; this is relevant in problems 7, 8, and parts of problems 5
and 8.
To integrate functions where the product rule is used to check the answer, we’ll later use
“integration by parts”; this is relevant in problems 4, and parts of problems 5 and 11.
An alternative to the quotient rule for differentiation is to write uv  uv 1 , and
differentiating using the product, then chain rule. Consequently, the method of substation
and integration by parts may be appropriate for integrating functions where the quotient
rule is used to check the answer. This is relevant in problem 10.
We’ll later see “partial fractions” is relevant in integrating the answers in problems 9 and
10.
We’ll later see additional integration formulas and techniques of integration; it’s unlikely
we would be asked to integrate such a function..
Integrating the answer from problem 12 would entail using a number of special tricks.
The general message is that corresponding to differentiation rules that one uses to check an
antiderivative answer are appropriate integration rules and / or integration techniques.
4